Question 3 of 6: Nanticoke Generating Station

Question 3: Nanticoke Generating Station (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 (from the Nanticoke technical specification on page 10):

Find. Part I: the flow and T–s diagrams, the four boundary enthalpies, and the boiler, cycle and overall efficiencies with an explanation of why the last two differ. Part II: the two expansion lines on the Mollier chart and the internal efficiency of the HP and the IP+LP turbines.

Approach. Read the four boundary enthalpies from the steam tables at the stated pressures and temperatures, sum the heat the working fluid absorbs in the main and reheat passes, and compare that in turn with the fuel heat and with the electrical output. For Part II, plot each expansion from its inlet state to its measured exit state and compare the actual enthalpy drop with the vertical (isentropic) drop to the same exit pressure.

1. Part I (a) — Diagrams. Figures 3.1 and 3.2 above carry the numbering used throughout: 1 = superheater outlet, 2 = cold reheat (HP exhaust), 3 = hot reheat, 4 = LP exhaust, 5 = condenser outlet, 6 = economiser inlet. The T–s diagram shows why reheat is used: the second expansion begins from 538 °C again, which raises the mean temperature of heat addition and simultaneously keeps the exhaust dry enough for the last-stage blading.
2. Part I (b) — Enthalpy of the water entering the boiler. State 6 is compressed liquid at 17.5 MPa and 252.5 °C. Saturation at 252.5 °C occurs at 4.15 MPa with \(h_f = 1097.6\ \text{kJ/kg}\); the compressed-liquid table (or the compressed-liquid correction) at 17.5 MPa gives essentially the same value, because for water at this temperature the pressure effect on enthalpy is almost cancelled by thermal expansion: $$h_6 \approx 1098\ \text{kJ/kg}$$ Note on method: the textbook shortcut \(h \approx h_f + v_f(p - p_{sat}) = 1097.6 + 0.001257(17\,500 - 4152) = 1114\ \text{kJ/kg}\) overshoots by 16 kJ/kg here because it ignores the \(-T(\partial v/\partial T)_p\) term; using it instead would change the boiler efficiency by only 0.5 percentage points.
3. Part I (b) concluded — Enthalpy of the superheated steam leaving the boiler. State 1 is superheated steam at 16.9 MPa and 538 °C. Interpolating the superheat tables, $$h_1 = 3396\ \text{kJ/kg},\qquad s_1 = 6.407\ \text{kJ/kg}\cdot\text{K}$$ $$\boxed{h_6 = 1098\ \text{kJ/kg},\qquad h_1 = 3396\ \text{kJ/kg}}$$
4. Part I (c) — Enthalpies of the reheat steam. The cold reheat entering the boiler is at 4.0 MPa and 343 °C, and the hot reheat leaving it is at 4.0 MPa and 538 °C: $$h_2 = 3076\ \text{kJ/kg},\qquad h_3 = 3533\ \text{kJ/kg}$$ $$\boxed{h_2 = 3076\ \text{kJ/kg (in)},\qquad h_3 = 3533\ \text{kJ/kg (out)}}$$ The reheater therefore adds 457 kJ/kg, about a fifth of what the main pass adds.
5. Part I (d) — Boiler thermal efficiency. The steam absorbs heat in two passes. Taking the reheat flow equal to the main steam flow (no extraction data are given), $$\dot Q_{main} = \dot m (h_1 - h_6) = 453.6 \times (3396 - 1098) = 1\,042\,400\ \text{kW}$$ $$\dot Q_{reheat} = \dot m (h_3 - h_2) = 453.6 \times (3533 - 3076) = 207\,400\ \text{kW}$$ $$\dot Q_{abs} = 1\,249\,800\ \text{kW}$$ The fuel supplies $$\dot Q_{fuel} = \dot m_{coal}\,CV = 47.9 \times 30\,240 = 1\,448\,500\ \text{kW}$$ $$\eta_{boiler} = \frac{1\,249\,800}{1\,448\,500}$$ $$\boxed{\eta_{boiler} = 0.863 = 86.3\ \%}$$ which is squarely in the 85–89 % band expected of a pulverised-coal boiler with an economiser and air heater.
6. Part I (e) — Cycle efficiency of the steam system. $$\eta_{cycle} = \frac{P_{gen}}{\dot Q_{abs}} = \frac{500\,000}{1\,249\,800}$$ $$\boxed{\eta_{cycle} = 0.400 = 40.0\ \%}$$ A 40 % gross cycle efficiency is exactly what a 16.9 MPa single-reheat unit of this era achieved.
7. Part I (f) — Overall plant efficiency. $$\eta_{overall} = \frac{P_{gen}}{\dot Q_{fuel}} = \frac{500\,000}{1\,448\,500}$$ $$\boxed{\eta_{overall} = 0.345 = 34.5\ \%}$$ Equivalently a heat rate of \(3600/0.3452 = 10\,430\ \text{kJ/kWh}\). As a check on the arithmetic, the three efficiencies must chain: \(0.863 \times 0.400 = 0.345\) ✓.
8. Part I (g) — Why (e) and (f) differ. They differ by exactly the boiler efficiency, because they measure the same electrical output against two different heat inputs. The cycle efficiency counts only the heat that actually crossed into the working fluid, so it grades the thermodynamic cycle — turbine quality, steam conditions, condenser vacuum, feed heating — and is limited fundamentally by the Carnot factor between the mean steam temperature and the condenser temperature. The overall efficiency counts the heat released by the coal, so it also carries every loss suffered before the water is reached: dry flue-gas loss up the stack (the largest, roughly 6 %), moisture from hydrogen in the fuel and from fuel moisture, unburnt carbon in ash, and radiation and convection from the boiler casing. Those losses amount to \(1\,448\,500 - 1\,249\,800 = 198\,700\ \text{kW}\), or 13.7 % of the fuel. The two are related by \(\eta_{overall} = \eta_{boiler}\times\eta_{cycle}\), and the distinction matters commercially: the boiler efficiency is the combustion engineer's figure of merit, the cycle efficiency is the turbine supplier's, and only the overall figure sets the fuel bill.
9. Part II (a) — Plotting the expansion lines. Figure 3.3 places both expansions on the Mollier (h–s) chart. The HP line runs from state 1 (16.9 MPa, 538 °C) to state 2 (4 MPa, 343 °C); the IP+LP line runs from state 3 (4 MPa, 538 °C) to state 4 on the 0.005 MPa isobar at \(x = 0.95\). Beside each is the vertical isentropic line from the same inlet state to the same exit pressure, whose lower end is 2s and 4s.
10. Part II (b) — Internal efficiency of the high-pressure turbine. Read off the chart: \(h_1 = 3396\), \(h_2 = 3076\), and the isentrope from state 1 meets 4 MPa at \(h_{2s} = 2987\ \text{kJ/kg}\). Then $$\eta_{HP} = \frac{h_1 - h_2}{h_1 - h_{2s}} = \frac{3396 - 3076}{3396 - 2987} = \frac{320}{409}$$ $$\boxed{\eta_{HP} = 0.78 = 78\ \%}$$
11. Part II (b) concluded — Internal efficiency of the low-pressure (IP + LP) turbine. The exhaust at 0.005 MPa with 5 % wetness has $$h_4 = h_f + x\,h_{fg} = 137.8 + 0.95 \times 2423.7 = 2440\ \text{kJ/kg}$$ and the isentrope from state 3 reaches the same pressure at \(h_{4s} = 2196\ \text{kJ/kg}\) (a much wetter \(x_{4s} = 0.85\)). Then $$\eta_{LP} = \frac{h_3 - h_4}{h_3 - h_{4s}} = \frac{3533 - 2440}{3533 - 2196} = \frac{1093}{1337}$$ $$\boxed{\eta_{LP} = 0.82 = 82\ \%}$$ Both are in the expected 78–85 % band, and the low-pressure machine is the better of the two — as it should be, since it has far more stages over which to recover the reheat factor, notwithstanding the moisture losses in its last rows. Chart readings carry perhaps ±1 percentage point.