Question 5 of 8: Power plant effluents

Question 5: Power plant effluents (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. Part I compares two 100 MW plants at 40 % and 50 % overall thermal efficiency burning carbon and methane respectively. Part II is a single 600 MW coal plant with the efficiency chain and the cooling-water data listed in the question.

Find. Part I: the combined-cycle carbon dioxide output as a percentage of the coal plant's, and a comment on the fuel assumptions. Part II: the cooling water mass and volume flows at the permitted temperature rise, the coal burn rate, and the ash made per day.

Approach. Part I converts electricity to fuel heat through the plant efficiency, fuel heat to fuel mass through the heating value, and fuel mass to carbon dioxide through the stoichiometry of the combustion equation. Part II builds the plant efficiency chain, then closes an energy balance on the steam cycle to find what the condenser must reject — which is the cycle heat minus the shaft work, not the fuel heat minus the electrical output.

1. Part I (a) — the stoichiometry of each fuel. Complete combustion of carbon and of methane gives one mole of carbon dioxide per mole of fuel in each case, but the fuel molar masses differ: $$\text{C} + \text{O}_2 \rightarrow \text{CO}_2: \qquad \frac{44}{12} = 3.667\ \text{kg CO}_2\ \text{per kg C}$$ $$\text{CH}_4 + 2\,\text{O}_2 \rightarrow \text{CO}_2 + 2\,\text{H}_2\text{O}: \qquad \frac{44}{16} = 2.750\ \text{kg CO}_2\ \text{per kg CH}_4$$ Methane already carries a quarter less carbon dioxide per kilogram of fuel, and it also has a much higher heating value, so the advantage per unit of heat is larger still.
2. Part I (a) — the coal fired plant. At 40 % overall efficiency, 100 MW of electricity needs $$\dot Q_{\text{coal}} = \frac{100\,000}{0.40} = 250\,000\ \text{kW}$$ $$\dot m_{\text{C}} = \frac{250\,000}{32\,800} = 7.622\ \text{kg/s}, \qquad \dot m_{\text{CO}_2} = 7.622 \times 3.667 = \boxed{27.95\ \text{kg/s}}$$ which is 100.6 t/h, or 1.006 kg of carbon dioxide per kilowatt-hour sent out — the familiar benchmark figure for a pure-carbon plant at this efficiency.
3. Part I (a) — the natural gas combined cycle. At 50 % the same output needs only 200 MW of fuel heat. Both efficiencies are taken on the same (higher heating value) basis for consistency: $$\dot Q_{\text{gas}} = \frac{100\,000}{0.50} = 200\,000\ \text{kW}, \qquad \dot m_{\text{CH}_4} = \frac{200\,000}{55\,530} = 3.602\ \text{kg/s}$$ $$\dot m_{\text{CO}_2} = 3.602 \times 2.750 = \boxed{9.90\ \text{kg/s}}$$ or 0.357 kg per kilowatt-hour.
4. Part I (a) — the comparison the question asks for. Expressing the gas plant as a percentage of the coal plant, $$\frac{\dot m_{\text{CO}_2,\text{gas}}}{\dot m_{\text{CO}_2,\text{coal}}} = \frac{9.90}{27.95} = \boxed{35.4\ \%}$$ so the substitution cuts carbon dioxide by 64.6 % for the same electricity. Roughly half of the saving comes from the fuel chemistry and half from the higher cycle efficiency. If the combined-cycle efficiency is instead quoted on the lower heating value, as is common practice for gas turbines, the fuel demand becomes 3.996 kg/s and the answer rises to 39.3 %; the conclusion is unchanged.
5. Part I (b) — how good are the two fuel assumptions? Neither is exact. A bituminous coal is typically 60–85 % carbon by mass as received, the balance being hydrogen (about 5 %), oxygen, nitrogen, sulphur, moisture and ash; the hydrogen releases heat without releasing any carbon dioxide, so a real coal emits slightly less carbon dioxide per unit of heat than pure carbon — the figure is about 0.34 kg CO₂ per kWh of fuel heat for pure carbon against roughly 0.32 for a typical bituminous coal, a 5 % overstatement in the coal figure. Pipeline natural gas is 87–96 % methane with ethane, propane and small amounts of carbon dioxide and nitrogen; the heavier hydrocarbons have a higher carbon-to-hydrogen ratio, so real gas emits a few per cent more carbon dioxide per unit of heat than pure methane. Both corrections push the same way: the true ratio would be a little higher than 35.4 %, perhaps 38–40 %, and the advantage of the combined cycle is slightly smaller than the idealised calculation suggests — but nothing like enough to change the engineering conclusion.
6. Part II — the plant efficiency chain. The three stated efficiencies multiply: $$\eta_{\text{overall}} = \eta_{\text{boiler}}\,\eta_{\text{cycle}}\,\eta_{\text{gen}} = 0.94 \times 0.41 \times 0.96 = 0.3700$$ $$\dot Q_{\text{fuel}} = \frac{600\,000}{0.3700} = 1\,621\,692\ \text{kW} = 1622\ \text{MW}$$ Working down that chain gives the two intermediate quantities the cooling water calculation needs: the heat that actually enters the steam cycle, $$\dot Q_{\text{cycle}} = 1\,621\,692 \times 0.94 = 1\,524\,390\ \text{kW}$$ and the shaft work the turbine has to produce to make 600 MW at the terminals, $$W_{\text{shaft}} = \frac{600\,000}{0.96} = 625\,000\ \text{kW}$$
7. Part II (a) — what the condenser must reject. The steam cycle is a closed loop, so everything that enters it as heat and does not leave as shaft work leaves through the condenser: $$\dot Q_{\text{rej}} = \dot Q_{\text{cycle}} - W_{\text{shaft}} = 1\,524\,390 - 625\,000 = \boxed{899\,390\ \text{kW}}$$ The check is immediate: $625\,000/1\,524\,390 = 0.410$, exactly the stated cycle efficiency. The two losses that are not in this figure matter: the 97 MW of boiler loss goes up the stack into the air, and the 25 MW of generator loss goes into the generator coolers, not into the circulating water.
8. Part II (a) — the cooling water flows. Holding the rise to the permitted 11 °C, $$\dot m_w = \frac{\dot Q_{\text{rej}}}{c_p\,\Delta t} = \frac{899\,390}{4.19 \times 11} = \boxed{19\,514\ \text{kg/s}}$$ $$\dot V_w = \frac{\dot m_w}{\rho} = \frac{19\,514}{1025} = \boxed{19.04\ \text{m}^3/\text{s}}$$ so the circulating water leaves at 24 °C. Nineteen cubic metres per second is a substantial river abstraction and explains why plants of this size are sited on large rivers, estuaries or the coast — or fitted with cooling towers.
9. Part II (b) — coal burn rate and ash production. The fuel heat and the as-received calorific value give the coal rate directly, and 6 % of it reports as ash: $$\dot m_{\text{coal}} = \frac{1\,621\,692}{35\,000} = \boxed{46.33\ \text{kg/s}}\ \ (166.8\ \text{t/h})$$ $$\dot m_{\text{ash}} = 0.06 \times 46.33 = 2.78\ \text{kg/s}, \qquad m_{\text{ash,day}} = 2.78 \times 86\,400 = \boxed{240.2\ \text{t/day}}$$ At roughly 80 % fly ash and 20 % bottom ash, that means about 190 t/day through the precipitator hoppers and 50 t/day out of the furnace bottom — four thousand tonnes a fortnight to be stored, sold to the cement industry or landfilled.
10. Guard against the classic short cut. A tempting but wrong route to part (a) is to say the condenser rejects the fuel heat minus the electrical output, $1\,621\,692 - 600\,000 = 1\,021\,692$ kW. That is 13.6 % too high, because it credits the condenser with the stack loss and the generator loss, neither of which ever touches the circulating water. It would size the cooling water system at 21.6 m³/s and, on a river-cooled station, would put the thermal discharge permit in the wrong place.