Question 3 of 8: Environmental Impact

Question 3: Environmental Impact (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. Part I: the coal burn rate and the carbon dioxide release rate, both in Mg/h. Part II: the circulating water flow, the tower's evaporative loss in three different units, and the make-up water each unit of electricity costs.

Approach. Part I is a single efficiency chain from electrical output back to coal, followed by a carbon balance on the combustion of that coal. Part II takes the condenser duty to the water side to get the circulating flow, reads the evaporative loss off the page-11 chart at the psychrometric state of the ambient air, and then expresses the same loss in the three units the question wants.

1. Part I (a) — the efficiency chain. The boiler converts fuel energy into steam energy and the cycle converts steam energy into electricity, so the two efficiencies multiply: $$\eta_{overall}=\eta_b\,\eta_{cy}=0.90\times 0.48=0.432 .$$ Working back from the generator terminals, $$\dot{Q}_{fuel}=\frac{\dot{P}_e}{\eta_{overall}}=\frac{600}{0.432} =1388.9\ \text{MJ/s}.$$
2. Part I (a) — coal burn rate. Dividing that heat demand by the calorific value of the coal as received, $$\dot{m}_{coal}=\frac{1388.9}{30}=46.30\ \text{kg/s},$$ and converting to the units asked for, $$\boxed{\dot{m}_{coal}=46.30\times\frac{3600}{1000}=166.7\ \text{Mg/h}} .$$ For context, that coal also carries 10 per cent ash, so the station makes $166.7\times 0.10=16.7$ Mg/h of ash — roughly 400 tonnes a day to be handled and disposed of.
3. Part I (b) — carbon balance. Every carbon atom in the fuel leaves as carbon dioxide when combustion is complete, and the molar masses convert mass of carbon to mass of carbon dioxide in the ratio 44:12: $$\dot{m}_{CO_2}=\dot{m}_{coal}\times x_C\times \frac{M_{CO_2}}{M_C} =46.30\times 0.75\times\frac{44}{12}=127.3\ \text{kg/s},$$ so that $$\boxed{\dot{m}_{CO_2}=458.3\ \text{Mg/h}} .$$ Referred to the electricity sent out this is $458\,300/600\,000=0.764$ kg of carbon dioxide per kilowatt-hour, at the low end of the roughly 0.75–1.0 kg/kWh range for coal units, as expected for a plant whose 43.2 per cent overall efficiency is in supercritical territory — a useful sanity check on the whole chain.

1. Part II (a) — circulating water flow. The condenser duty is carried away by the cooling water through its 10 K rise: $$\dot{m}_w=\frac{\dot{Q}_{rej}}{c_{p,w}\,\Delta T_w} =\frac{1\,500\,000}{4.19\times 10}=35\,800\ \text{kg/s},$$ which at 1000 kg/m³ is $$\boxed{\dot{V}_w=35.8\ \text{m}^3\text{/s}} .$$ This is the flow circulating between condenser and tower; only a small fraction of it is actually consumed.
2. Part II (b) — entering the page-11 chart. The chart is entered on dry bulb horizontally and wet bulb vertically, so the ambient state must first be converted. Air at 30 °C dry bulb and 40 per cent relative humidity has a thermodynamic wet-bulb temperature of about 20 °C, and the point (30, 20) falls on the chart's own 40 per cent humidity line, which is the confirmation that the entry point is right.
   Figure 3.1 — the page-11 evaporative-loss chart with the operating point marked. Dash-dot lines are constant relative humidity; heavy lines are contours of constant evaporative loss in m³/GJ rejected. Entering at 30 °C dry bulb and 20 °C wet bulb lands on the 0.375 contour. Note that several contours are labelled at both ends and the label sits to the right of the line it belongs to, so the value must be traced along the line rather than read from the nearest label.
3. Part II (b) — evaporative loss in the two units asked for. Reading the chart at that point gives $$\boxed{C_{evap}=0.375\ \text{m}^3\text{/GJ rejected} =3.75\times 10^{-4}\ \text{m}^3\text{/MJ rejected}} ,$$ and multiplying by the tower's actual duty of 1.5 GJ/s, $$\boxed{\dot{V}_{evap}=0.375\times 1.5=0.5625\ \text{m}^3\text{/s}} .$$ The reading can be bounded independently: if the tower rejected its heat entirely as latent heat at 30 °C the loss would be $1\,000\,000/2430.5=411$ kg per GJ, i.e. 0.411 m³/GJ. The chart value is 91 per cent of that ceiling, so 9 per cent of the duty leaves as sensible heating of the air — exactly what one expects of warm, fairly dry air, and a reading above 0.411 would have been impossible.
4. Part II (c) — loss as a fraction of the circulating flow. Comparing the two flows already found, $$\boxed{\frac{\dot{V}_{evap}}{\dot{V}_w} =\frac{0.5625}{35.8}=0.0157=1.57\ \text{per cent}} .$$ This is the classic result that a wet cooling tower evaporates between one and two per cent of its circulating flow, and it is what sets the make-up duty; the remaining losses (drift and the blowdown needed to control the cycles of concentration) are additional but smaller.
5. Part II (d) — water consumed per unit generated. Referring the evaporative loss to the electricity sent out, $$\boxed{\frac{\dot{V}_{evap}}{\dot{P}_e} =\frac{0.5625\times 1000\times 3600}{600\,000}=3.375\ \text{L/kWh}} .$$ For a 600 MW unit at full load this is about 2.0 megalitres an hour, or 49 megalitres a day — roughly the residential supply of a city of 170 000 people at about 290 litres per person per day — which is why water availability, not fuel, often decides where a thermal station can be sited in Canada's drier interior.