Question 4 of 8: Steam Plant Design

Question 4: Steam Plant Design (10 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.

Find. Annual energy production at both the actual and the maximum rate; the annual coal tonnage and the resulting train movements per day; and the three annual cost streams, each expressed both in dollars per year and in cents per kilowatt-hour, together with their total.

Approach. Energy first, because every later quantity is a rate multiplied by it. The heat rate is stated for the whole plant on a fuel basis, so it converts electricity directly into coal without any further use of the boiler efficiency; the two capital-derived costs are simple percentages of a capital sum; and each cost per unit is an annual cost divided by the annual actual production.

1. Part (a) — annual energy production. A plant at its maximum continuous rating for every hour of the year would deliver $$E_{max}=500\,000\ \text{kW}\times 8760\ \text{h}=4.380\times 10^{9}\ \text{kWh},$$ and the capacity factor scales that to what the plant actually sends out: $$\boxed{E_{act}=0.80\times 4.380\times 10^{9}=3.504\times 10^{9}\ \text{kWh}} .$$ Only $E_{act}$ earns revenue, so it is the denominator of every unit cost below, while $E_{max}$ is what the capital was bought to be able to produce.
2. Part (b) — annual coal requirement. The heat rate is the fuel energy needed per unit of electricity sent out, so the annual fuel demand and hence the coal tonnage follow directly: $$m_{coal}=\frac{E_{act}\times HR}{CV} =\frac{3.504\times 10^{9}\times 10\,550}{24\,000}=1.5403\times 10^{9}\ \text{kg},$$ $$\boxed{m_{coal}=1.540\times 10^{6}\ \text{Mg per year}} .$$ Note that the 90 per cent boiler efficiency is not applied here: the heat rate is already a whole-plant, fuel-based figure, and applying the boiler efficiency a second time would double-count the stack loss.
3. Part (b) — train movements. Spreading the annual tonnage evenly over the year gives $1.5403\times 10^{6}/365=4220$ Mg per day, and one train carries $50\times 60=3000$ Mg, so $$\boxed{n_{trains}=\frac{4220}{3000}=1.41\ \text{trains per day}} .$$ In practice this means a unit train roughly every seventeen hours, or ten trains a week — the number that sizes the rail loop, the tippler and the stockyard.
4. Part (c) — cost of coal. At the quoted price, $$C_{coal}=1.5403\times 10^{6}\times 100=154.0\ \text{million CAD per year},$$ and dividing by the electricity actually sent out, $$\boxed{c_{coal}=\frac{154.03\times 10^{6}}{3.504\times 10^{9}}\times 100 =4.40\ \text{cent/kWh}} .$$
5. Part (d) — capital repayment. The capital sum is $2500\times 500\,000=1.250\times 10^{9}$ CAD, of which 10 per cent is repaid annually: $$C_{cap}=0.10\times 1.250\times 10^{9}=125.0\ \text{million CAD per year},$$ $$\boxed{c_{cap}=\frac{125.0\times 10^{6}}{3.504\times 10^{9}}\times 100 =3.57\ \text{cent/kWh}} .$$ This is a fixed charge: it is incurred whether the plant runs or not, which is why the capacity factor matters so much to the unit cost.
6. Part (e) — administration and maintenance. Eight per cent of the same capital sum, $$C_{adm}=0.08\times 1.250\times 10^{9}=100.0\ \text{million CAD per year},$$ $$\boxed{c_{adm}=\frac{100.0\times 10^{6}}{3.504\times 10^{9}}\times 100 =2.85\ \text{cent/kWh}} .$$
7. Part (f) — total production cost. Adding the three streams, $$\boxed{c_{total}=4.40+3.57+2.85=10.82\ \text{cent/kWh}} .$$ Fuel is 40.6 per cent of the total and the two capital-derived charges together are 59.4 per cent, so this is a capital-dominated plant — the opposite of a simple-cycle peaking gas turbine, and the reason coal units are run base-load.
   Figure 4.1 — Part (f): the production cost broken into its three streams, cent/kWh. Fuel is the largest single item but the two capital-derived charges together dominate, which is what makes the capacity factor the most sensitive assumption in the whole estimate.