Question 2 of 6: Steam Plant Design

Question 2: Steam Plant Design (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. A proposed 500 MW coal-fired station, characterised by its whole-plant heat rate, boiler efficiency, capital cost and fuel properties, to be costed and resourced over a year of operation at a capacity factor of 0.80.

Find. The annual energy production, coal tonnage and train movements, the three components of the unit generating cost and their total, and the full-load condenser duty with the cooling-water flow it demands.

Approach. Work the energy chain forwards — hours to kWh, kWh to fuel through the heat rate, fuel to tonnes and dollars — then divide each annual cost by the same annual production to get comparable cent/kWh figures; finally close a full-load energy balance on the steam cycle only, not on the fuel.

1. Part (a) — annual production. A year contains 8760 hours, so the maximum possible output is the nameplate rating run flat out for all of them, and the actual output is that reduced by the capacity factor:$$E_{max}=500\,000\ \text{kW}\times 8760\ \text{h}=4.380\times 10^{9}\ \text{kWh}$$$$\boxed{E_{act}=0.80\times 4.380\times 10^{9}=3.504\times 10^{9}\ \text{kWh/yr}}$$Everything that follows is divided by $E_{act}$, never by $E_{max}$: the station earns revenue only on the energy it actually sends out.
2. Part (b) — coal tonnage and train movements. The heat rate converts sent-out energy straight into fuel heat, and the heating value converts that into mass:$$m_{coal}=\frac{E_{act}\times \text{HR}}{\text{HV}}=\frac{3.504\times 10^{9}\times 10\,550}{24\,000\times 10^{3}}\qquad\Rightarrow\qquad \boxed{m_{coal}=1.540\times 10^{6}\ \text{Mg/yr}}$$That is 4220 Mg per day. One train carries $50\times 60 = 3000$ Mg, so$$\boxed{N_{trains}=\frac{4220}{3000}=1.41\ \text{trains per day}}$$In practice this is scheduled as ten unit trains a week. A 30-day live stockpile at this burn rate is about 127 000 Mg of coal, which sets the size of the yard.
3. Part (c) — the fuel cost. At CAD 100 per Mg,$$C_{coal}=1.540\times 10^{6}\times 100=\text{CAD}\ 154.0\ \text{million per year}$$$$\boxed{c_{coal}=\frac{154.03\times 10^{6}}{3.504\times 10^{9}}\times 100=4.40\ \text{cent/kWh}}$$The factor of 100 turns dollars per kWh into cents per kWh; forgetting it is the commonest slip in this question.
4. Part (d) — capital repayment. The plant costs $2500 \times 500\,000 = $ CAD 1.250 billion, of which 10 % is repaid annually:$$C_{cap}=0.10\times 1.250\times 10^{9}=\text{CAD}\ 125.0\ \text{million per year}$$$$\boxed{c_{cap}=\frac{125.0\times 10^{6}}{3.504\times 10^{9}}\times 100=3.57\ \text{cent/kWh}}$$Note that the 40-year life does not enter the arithmetic: the question fixes the repayment as a flat percentage rather than asking for a capital-recovery factor. The life expectancy matters instead as a reasonableness test — ten per cent a year repays the capital, with interest, comfortably inside the plant's life.
5. Part (e) — administration and maintenance. The same base at 8 %:$$C_{adm}=0.08\times 1.250\times 10^{9}=\text{CAD}\ 100.0\ \text{million per year}$$$$\boxed{c_{adm}=\frac{100.0\times 10^{6}}{3.504\times 10^{9}}\times 100=2.85\ \text{cent/kWh}}$$Basing operating cost on capital rather than on output is a crude but common screening convention; it reflects the fact that most of a thermal station's maintenance bill is fixed by the size of the plant, not by how hard it is run.
6. Part (f) — the total unit cost. The three components simply add:$$\boxed{c_{total}=4.40+3.57+2.85=10.82\ \text{cent/kWh}}$$Fuel is 40.6 % of the total and the two capital-related items together are 59.4 %. That split is the single most useful output of the calculation: it says the station is capital-dominated, so its economics depend far more on being run at a high capacity factor than on the price of coal.
7. Part (g) — heat rejected at full load. This is the part that separates a correct answer from a plausible one. First the fuel heat input at the full 500 MW:$$\dot{Q}_{fuel}=\frac{P_e\times \text{HR}}{3600}=\frac{500\,000\times 10\,550}{3600}=1\,465\,278\ \text{kJ/s}$$Only the fraction that crosses the boiler surfaces enters the steam cycle; the rest goes up the stack:$$\dot{Q}_{cycle}=\eta_b\dot{Q}_{fuel}=0.90\times 1\,465\,278=1\,318\,750\ \text{kJ/s}$$and the condenser receives what the cycle does not turn into work:$$\boxed{\dot{Q}_{rej}=\dot{Q}_{cycle}-P_e=1\,318\,750-500\,000=818\,750\ \text{kJ/s}}$$Using $\dot{Q}_{fuel}-P_e$ instead would give 965 278 kJ/s and overstate the condenser by 17.9 %, because it credits the condenser with the 146 528 kJ/s that left up the chimney. The turbine-hall cycle efficiency implied here is $500\,000/1\,318\,750 = 37.9$ %, against a whole-plant 34.1 %, and the difference between those two figures is exactly the boiler efficiency.
8. Part (h) — cooling water. With a 10 °C rise and the constants sheet value $c_p = 4.19$ kJ/kg·K,$$\dot{m}_w=\frac{\dot{Q}_{rej}}{c_p\,\Delta T}=\frac{818\,750}{4.19\times 10}=19\,541\ \text{kg/s}$$$$\boxed{\dot{V}_w=\frac{19\,541}{1000}=19.5\ \text{m}^3\text{/s}}$$For scale, that is roughly the mean flow of a small river, and it is why a station of this size is sited on a large water body or given cooling towers.