Question 1 of 6: Plant Energy Balances

Question 1: Plant Energy Balances (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.

Part I — Combined cycle configuration (9 marks)

Given.

Find. The combined thermal efficiency of the plant, and the capacity (MW) each of the two turbines must have, together with a flow diagram carrying every heat flow rate in MJ/s and every power output in MW.

Approach. Follow one unit of fuel heat through the plant in series: the gas turbine takes its share as work, the heat recovery boiler captures a fraction of what leaves in the exhaust, and the steam cycle converts a share of that — so the two cycle efficiencies multiply rather than add on the recovered stream.

1. Part (a) — write the combined efficiency as a series cascade. Of the fuel heat \(\dot Q_f\), the gas turbine delivers \(\eta_{gt}\dot Q_f\) as work and rejects the remaining \((1-\eta_{gt})\dot Q_f\) in its exhaust. The heat recovery boiler transfers a fraction \(\eta_{hrb}\) of that exhaust stream into the steam, and the steam cycle converts a fraction \(\eta_{st}\) of what it receives: \[ \eta_{cc} \;=\; \eta_{gt} \;+\; (1-\eta_{gt})\,\eta_{hrb}\,\eta_{st} \]
2. Substitute the given efficiencies. \[ \eta_{cc} = 0.30 + (1-0.30)(0.90)(0.30) = 0.30 + 0.189 \] \[ \boxed{\eta_{cc} = 0.489 \;\;\text{(48.9\%)}} \] The 18.9 percentage points contributed by the bottoming cycle are what makes a combined cycle worth building: neither cycle alone reaches 49%.
3. Size the fuel heat input from the required electrical output. With the plant efficiency known, the fuel rate follows directly: \[ \dot Q_f = \frac{P_{tot}}{\eta_{cc}} = \frac{250\ \text{MW}}{0.489} = 511.25\ \text{MJ/s} \]
4. Part (b) — gas turbine capacity. The topping cycle takes its own efficiency on the whole fuel stream: \[ P_{gt} = \eta_{gt}\,\dot Q_f = 0.30 \times 511.25 \] \[ \boxed{P_{gt} = 153.4\ \text{MW}} \]
5. Part (c) — steam turbine capacity. The bottoming cycle sees only the recovered heat, \(0.70 \times 0.90 = 0.63\) of the fuel input: \[ P_{st} = (1-\eta_{gt})\,\eta_{hrb}\,\eta_{st}\,\dot Q_f = 0.189 \times 511.25 \] \[ \boxed{P_{st} = 96.6\ \text{MW}} \] Checking the split, \(153.4 + 96.6 = 250.0\) MW, so the two capacities re-add to the specified plant output. The roughly 61:39 division between topping and bottoming cycle is characteristic of single-pressure heat recovery plant.
6. Compute the heat flow rates annotated on the sketch. Each is a fixed fraction of \(\dot Q_f\): the gas turbine exhaust carries \(0.70\dot Q_f = 357.9\) MJ/s, of which the boiler passes \(0.63\dot Q_f = 322.1\) MJ/s into the steam and loses \(0.07\dot Q_f = 35.8\) MJ/s up the stack. The steam cycle rejects the difference between what it takes in and what it makes: \[ \dot Q_{cond} = 322.1 - 96.6 = 225.5\ \text{MJ/s} \] The plant balance closes: \(153.4 + 96.6 + 35.8 + 225.5 = 511.3\) MJ/s in, which is \(\dot Q_f\).

Part II — Carbon dioxide emissions (6 marks)

Given. Coal treated as pure carbon, HHV = LHV = 32 800 kJ/kg, burnt in a plant of thermal efficiency 40%. Natural gas treated as pure methane, HHV = 55 530 kJ/kg and LHV = 50 050 kJ/kg, burnt in a combined cycle of thermal efficiency 50%. Both plants are to deliver the same electrical energy.

Find. The mass of CO₂ released by the gas plant expressed as a percentage of that released by the coal plant for equal electricity sent out, plus an assessment of the two purity assumptions.

Approach. Work per unit of electrical output. Divide by the plant efficiency to get the fuel heat needed, divide by the heating value to get fuel mass, then multiply by the stoichiometric CO₂ yield of the combustion equation.

1. Write the two combustion equations and their CO₂ yields. Carbon burns to \[ \mathrm{C} + \mathrm{O_2} \rightarrow \mathrm{CO_2} \] so 12 kg of carbon yields 44 kg of CO₂, a mass ratio of \(44/12 = 3.667\). Methane burns to \[ \mathrm{CH_4} + 2\,\mathrm{O_2} \rightarrow \mathrm{CO_2} + 2\,\mathrm{H_2O} \] so 16 kg of methane yields 44 kg of CO₂, a ratio of \(44/16 = 2.750\). Methane's advantage begins here: a quarter of its combustion energy comes from burning hydrogen to water, which produces no CO₂ at all.
2. Form the specific emission per unit of electricity. For a fuel of heating value \(HV\) burnt at plant efficiency \(\eta\), \[ \frac{m_{CO_2}}{E_{elec}} = \frac{1}{\eta\,HV}\cdot\frac{M_{CO_2}}{M_{fuel}} \] The heating value must be taken on the same basis for both fuels, since the plant efficiencies are quoted on one basis; the higher heating value is the North American convention and is used for the primary answer.
3. Evaluate for coal. Fuel heat needed per kJ of electricity is \(1/0.40 = 2.50\) kJ, so the carbon burnt is \(2.50/32\,800 = 7.622\times10^{-5}\) kg per kJ_e, and \[ \frac{m_{CO_2}}{E} \bigg|_{coal} = 7.622\times10^{-5} \times 3.667 = 2.795\times10^{-4}\ \text{kg/kJ}_e = 0.2795\ \text{kg/MJ}_e \]
4. Evaluate for the gas combined cycle. Fuel heat needed is \(1/0.50 = 2.00\) kJ per kJ_e, the methane burnt is \(2.00/55\,530 = 3.602\times10^{-5}\) kg, and \[ \frac{m_{CO_2}}{E} \bigg|_{gas} = 3.602\times10^{-5} \times 2.750 = 9.905\times10^{-5}\ \text{kg/kJ}_e = 0.09905\ \text{kg/MJ}_e \]
5. Take the ratio. \[ \frac{m_{CO_2,gas}}{m_{CO_2,coal}} = \frac{9.905\times10^{-5}}{2.795\times10^{-4}} \] \[ \boxed{\text{Gas produces } 35.4\% \text{ of the CO}_2 \text{ of coal — a 64.6\% reduction}} \] Repeating the calculation on the lower heating value of methane (50 050 kJ/kg, with carbon unchanged because it contains no hydrogen) gives 39.3%. The answer is therefore 35–39% depending on the heating-value basis, and roughly two-thirds of the emission is avoided either way. About half the saving comes from the fuel chemistry and half from the 40% → 50% efficiency gain.
6. Comment on the validity of the two purity assumptions. Neither is strictly valid, and they err in opposite directions, so the calculated 35% understates the true ratio.

   Coal is not pure carbon. A Canadian bituminous or sub-bituminous coal is typically 65–80% carbon by mass, with some 4–5% hydrogen plus sulphur, nitrogen, oxygen, inherent moisture and 8–15% ash. Because the hydrogen releases heat without releasing CO₂, real coal carries less carbon per unit of heat than pure carbon does: pure carbon is \(3.667/32\,800 = 111.8\) kg CO₂/GJ, whereas measured factors for real bituminous coal are near 94 kg CO₂/GJ. The pure-carbon idealisation therefore overstates coal's emission by roughly 18%.

   Natural gas is not pure methane either. Pipeline gas typically carries 3–8% ethane and heavier hydrocarbons plus some CO₂ and nitrogen. Higher hydrocarbons have a higher carbon-to-hydrogen ratio than methane, so real gas carries slightly more carbon per unit of heat: about 50.6 kg CO₂/GJ (HHV) against 49.5 for pure methane.

   Substituting those measured factors while keeping the given plant efficiencies, \[ \frac{50.6/0.50}{94.6/0.40} = 42.8\% \] so the honest answer for real fuels is about 43% rather than 35%. The qualitative conclusion is unchanged and remains robust — switching from coal to a gas combined cycle removes over half the CO₂ per unit of electricity — but the idealisation flatters the switch by some eight percentage points. A complete life-cycle comparison would also have to price upstream methane leakage, since methane is a far more potent greenhouse gas than CO₂ over a twenty-year horizon; a leakage rate of a few percent erodes a noticeable part of the benefit computed here.