22-Mec-B3 Energy Conversion and Power Generation · May 2014

Question 1 of 6: Steam Plant Design

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Question 1: 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.

Find. The annual energy, fuel and cost budget of the station on a cent/kWh basis, and the full-load condenser duty with the circulating-water flow it demands.

Approach. Convert the capacity and capacity factor into annual kilowatt-hours, run the heat rate backwards to get fuel energy and hence coal tonnage, divide each annual cost by the annual output to obtain a levelised cent/kWh, and close with a full-load energy balance across the turbine hall alone.

1. Part (a) — annual maximum and actual electrical production. The maximum possible production is the nameplate capacity held for every hour of the year, and the actual production is that figure scaled by the capacity factor: $$E_{\max} = P_e \times 8760 = 500\,000\ \text{kW} \times 8760\ \text{h} = 4.380 \times 10^{9}\ \text{kWh}$$ $$E_{\text{act}} = CF \times E_{\max} = 0.80 \times 4.380 \times 10^{9} = \boxed{3.504 \times 10^{9}\ \text{kWh/yr}}$$ The capacity factor absorbs everything that keeps a unit off bars — planned outages, forced outages, and running below maximum continuous rating when the system does not need the output.
2. Part (b) — annual coal tonnage. The heat rate is the fuel energy charged per kilowatt-hour sent out, so the annual fuel energy follows directly and the coal mass is that energy divided by the heating value: $$Q_{\text{fuel}} = E_{\text{act}} \times HR = 3.504 \times 10^{9} \times 10\,550 = 3.6967 \times 10^{13}\ \text{kJ/yr}$$ $$m_{\text{coal}} = \frac{Q_{\text{fuel}}}{CV} = \frac{3.6967 \times 10^{13}}{24\,000} = 1.5403 \times 10^{9}\ \text{kg} = \boxed{1.540 \times 10^{6}\ \text{Mg/yr}}$$ Note that the boiler efficiency is not applied here. A heat rate quoted for the "whole plant" is already the fuel-to-busbar figure; applying the 90 % a second time would double-count the furnace loss.
3. Part (b) continued — unit trains per day. Averaged over the year the station burns $$\dot m_{\text{coal}} = \frac{1.540 \times 10^{6}}{365} = 4\,220\ \text{Mg/day}$$ and one unit train carries $60 \times 50 = 3\,000$ Mg, so $$N_{\text{trains}} = \frac{4\,220}{3\,000} = \boxed{1.41\ \text{trains/day}}$$ In practice this is scheduled as ten trains a week. The coal yard must hold enough for the winter peak plus a strike or freeze-up reserve, so a 30-day live pile of roughly 130 000 Mg is typical.
4. Part (c) — annual coal cost and cost per unit sent out. At 100 $/Mg, $$C_{\text{coal}} = 1.5403 \times 10^{6}\ \text{Mg} \times 100\ \tfrac{\$}{\text{Mg}} = \boxed{\$154.0\ \text{million per year}}$$ Dividing by the annual output, or equivalently working straight from the heat rate, $$c_{\text{coal}} = \frac{HR}{CV} \times \text{price} = \frac{10\,550}{24\,000}\ \tfrac{\text{kg}}{\text{kWh}} \times 0.100\ \tfrac{\$}{\text{kg}} = 0.04396\ \tfrac{\$}{\text{kWh}} = \boxed{4.40\ \text{cent/kWh}}$$ The second route is worth knowing because it is independent of capacity factor: fuel cost per kilowatt-hour depends only on efficiency and price, which is exactly why fuel is treated as the variable cost in economic dispatch.
5. Part (d) — capital repayment. The overnight capital cost is $$C_{\text{cap}} = 2\,500\ \tfrac{\$}{\text{kW}} \times 500\,000\ \text{kW} = \$1.250 \times 10^{9}$$ so the annual repayment charge is $$A_{\text{cap}} = 0.10 \times 1.250 \times 10^{9} = \boxed{\$125.0\ \text{million per year}}$$ $$c_{\text{cap}} = \frac{125.0 \times 10^{6}}{3.504 \times 10^{9}} \times 100 = \boxed{3.57\ \text{cent/kWh}}$$ Unlike fuel, this charge is fixed: it is incurred whether the unit runs or not, so it falls per kilowatt-hour as the capacity factor rises. That asymmetry is the reason capital-intensive plant is base-loaded.
6. Part (e) — administration and maintenance. By the same route at 8 %, $$A_{\text{adm}} = 0.08 \times 1.250 \times 10^{9} = \boxed{\$100.0\ \text{million per year}}$$ $$c_{\text{adm}} = \frac{100.0 \times 10^{6}}{3.504 \times 10^{9}} \times 100 = \boxed{2.85\ \text{cent/kWh}}$$
7. Part (f) — total production cost. Adding the three components, $$c_{\text{total}} = 4.396 + 3.567 + 2.854 = \boxed{10.82\ \text{cent/kWh}}$$ Fuel is 41 % of the total and the two fixed charges together are 59 %. A useful check on the shape of the answer: the stated 40-year life and 10 % annual repayment imply a simple payback of ten years, which is a much heavier capital charge than a 40-year annuity at any plausible discount rate would produce, so this figure is conservative by design.
8. Part (g) — full-load heat rejection to the cooling water. At full load the station sends out 500 MW, so the fuel energy rate is $$\dot Q_{\text{fuel}} = P_e \times \frac{HR}{3600} = 500\,000 \times \frac{10\,550}{3600} = 1\,465\,278\ \text{kJ/s}$$ Only the fraction that crosses the boiler tubes reaches the working fluid: $$\dot Q_{\text{cycle}} = \eta_b \dot Q_{\text{fuel}} = 0.90 \times 1\,465\,278 = 1\,318\,750\ \text{kJ/s}$$ The steam cycle is a closed control volume that receives $\dot Q_{\text{cycle}}$ and delivers the electrical output, so by the first law everything else leaves in the circulating water: $$\dot Q_{\text{rej}} = \dot Q_{\text{cycle}} - P_e = 1\,318\,750 - 500\,000 = \boxed{818\,750\ \text{kJ/s}}$$ This is the single most common place to slip. Writing $\dot Q_{\text{rej}} = \dot Q_{\text{fuel}} - P_e$ would give 965 278 kJ/s and overstate the cooling water by 18 %, because it debits the condenser with the stack loss that in fact leaves through the chimney. The corresponding turbine-hall efficiency, $500\,000/1\,318\,750 = 37.9\ \%$, is the right order for a subcritical reheat cycle, whereas the whole-plant efficiency is $3600/10\,550 = 34.1\ \%$.
9. Part (h) — circulating-water quantity. With a 10 °C rise and the paper's $c_p = 4.190\ \text{kJ/kg}\cdot\text{K}$, $$\dot m_w = \frac{\dot Q_{\text{rej}}}{c_p \Delta T_w} = \frac{818\,750}{4.190 \times 10} = 19\,541\ \text{kg/s}$$ $$\dot V_w = \frac{\dot m_w}{\rho} = \frac{19\,541}{1000} = \boxed{19.54\ \text{m}^3\text{/s}}$$ That is roughly 70 000 m³ per hour, which for a once-through station means a river or tidal intake; on an inland site the same duty needs two natural-draught towers of about 100 m and an evaporative make-up of some 2 % of the circulating flow.