22-Mec-B3 Energy Conversion and Power Generation · Undated paper
Question 5 of 8: Wind and Water Power
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Examinations, May 2019 — 16-Mec-B3 Energy Conversion and Power Generation. Three hours, closed book. Section A is calculative (Questions 1–5) and Section B descriptive (Questions 6–8); candidates answer four questions from Section A and two from Section B, six questions of ten marks each for a total of sixty. Reference data for particular questions are bound in as pages 10–17, reference formulae and constants as pages 18–21, and steam tables from Thermodynamics and Heat Power are supplied. All eight questions are solved here.
Reference texts.
Granet & Bluestein, Thermodynamics and Heat Power, 6th ed. — steam tables, vapour and gas power cycles (the tables bound into this paper).
El-Wakil, Powerplant Technology — steam generators, gas-turbine and combined-cycle plant, wind and hydro conversion, plant siting and environmental impact.
Çengel & Boles, Thermodynamics: An Engineering Approach, 9th ed. — regenerative Rankine cycles, Brayton-cycle modifications, isentropic and internal efficiencies.
Çengel & Ghajar, Heat and Mass Transfer — heat-recovery steam generator temperature profiles and pinch analysis.
Manwell, McGowan & Rogers, Wind Energy Explained, 2nd ed. — actuator-disc theory, the Betz limit and power coefficients.
Natural Resources Canada, Canadian Renewable Energy Atlas, and CNSC REGDOC series — Canadian generation mix, siting and licensing context for Question 8.
The combustion balance of Question 1 returns a gas mass flow of 125.4 kg/s, which matches the 125 kg/s that Question 2 states. Readings taken from printed charts are identified explicitly wherever they occur.
Check: water and steam properties used below are IAPWS values, the formulation the bound Granet & Bluestein tables tabulate; every reading agrees with those tables to better than 0.1 %, comfortably inside the paper’s own rounding. Where a value had to be read off a printed chart (the Page 13 power curve and the Page 14 efficiency curves) the reading is stated explicitly and carries roughly ±1 % of graph-reading uncertainty. Every boxed result is recomputed from the question’s own data for this paper.
Given. Part I — a three-bladed horizontal-axis machine of 80 m diameter turning at 15.7 rev/min in a 10 m/s wind. Part II — a Kaplan unit with pressure and velocity measurements at two sections.
Question 5 — given data
Quantity
Symbol
Value
Rotor diameter
$D$
80 m
Swept area
$A$
5027 m²
Rotational speed
$N$
15.7 rev/min
Wind speed
$V_1$
10 m/s
Air density (power-curve basis)
$\rho_a$
1.225 kg/m³
Nominal output
—
1.8 MW at 16 m/s
Hydro flow rate
$Q$
354 m³/s
Generator output
$P_e$
110 MW
Inlet / outlet diameters
$D_1,\ D_2$
6.4 m, 7.0 m
Inlet pressure
$p_1$
226 kPa gauge
Outlet pressure
$p_2$
−4.5 m H₂O
Elevation drop, inlet to outlet
$\Delta z$
5.0 m
Find. Part I: the Betz power and efficiency, the ideal and actual powers read from the Page 14 chart at the machine’s own tip-speed ratio, and the manufacturer’s output at 10 m/s. Part II: the overall turbine-generator efficiency at Mactaquac.
Question 5 Part I — the actuator-disc stream tube. The rotor can only extract power by slowing the flow, and slowing it too much chokes the mass flow through the disc; the optimum is the Betz condition.
Approach. Part I: compute the kinetic power crossing the swept area, then scale it by the Betz factor and by the two efficiencies read from the chart at the tip-speed ratio the machine actually runs at. Part II: apply the steady-flow energy equation between the two measuring sections to get the head, and compare the hydraulic power with the electrical output.
Part I(a) — the power in the wind. The kinetic energy flux through the swept area is $$P_{wind} = \tfrac{1}{2}\rho_a A V_1^{3} = 0.5 \times 1.225 \times 5027 \times 10^{3} = 3079\ \text{kW}.$$ This is the reference against which every efficiency below is quoted.
Part I(a) — Betz limit. The reference equation on Page 21 gives the maximum ideal power directly: $$P_{max} = \frac{8\rho_a A V_1^{3}}{27} = \frac{8 \times 1.225 \times 5027 \times 1000}{27}$$ $$\boxed{P_{max} = 1825\ \text{kW},\qquad \eta_{max} = \frac{16}{27} = 59.3\ \%}$$ No axial-flow rotor of any design can beat this at any wind speed.
Part I(b) — the operating tip-speed ratio. The blade tip travels at $$U = \frac{\pi D N}{60} = \frac{\pi \times 80 \times 15.7}{60} = 65.8\ \text{m/s},$$ so the ratio of blade tip speed to wind speed is $$\lambda = \frac{U}{V_1} = \frac{65.8}{10} = 6.58.$$ This is the abscissa at which the Page 14 chart must be read.
Part I(b) — ideal power from the chart. At $\lambda = 6.58$ the curve marked ideal efficiency for propeller-type windmills reads 58 %, just short of its 59.3 % asymptote. Hence $$\boxed{\eta_{ideal} = 58\ \%,\qquad P_{ideal} = 0.58 \times 3079 = 1786\ \text{kW}}$$ The ideal curve falls below the Betz value at low $\lambda$ because a slowly turning rotor leaves swirl in the wake; by $\lambda \approx 6$ that loss is nearly gone, which is precisely why modern machines are designed to run here.
Part I(c) — actual power from the chart. At the same $\lambda = 6.58$ the real-machine curve on Page 14 — the high-speed propeller-type characteristic, the only measured curve that extends to this tip-speed ratio — reads 44 %: $$\boxed{\eta_{actual} = 44\ \%,\qquad P_{actual} = 0.44 \times 3079 = 1355\ \text{kW}}$$ The 14-point gap from the ideal curve is blade drag, tip loss and the finite number of blades.
Part I(d) — the manufacturer’s figure. Reading the V80 power curve on Page 13 at 10 m/s gives an electrical output of $$\boxed{P_{spec} \approx 1150\ \text{kW}}$$ an overall efficiency of $1150/3079 = 37$ % from wind to terminals. The comparison the question asks for therefore runs 59.3 % (Betz) → 58 % (ideal propeller) → 44 % (real rotor) → 37 % (delivered), the last step being gearbox, generator and converter losses of roughly 15 % together with the pitch control that begins to shed power as the machine approaches its 1.8 MW rating at 16 m/s.
Part II — velocities at the two measuring sections. With the same volume flow through both pipes, $$A_1 = \frac{\pi (6.4)^2}{4} = 32.17\ \text{m}^2,\qquad A_2 = \frac{\pi (7.0)^2}{4} = 38.49\ \text{m}^2,$$ $$V_1 = \frac{354}{32.17} = 11.00\ \text{m/s},\qquad V_2 = \frac{354}{38.49} = 9.20\ \text{m/s}.$$
Part II — net head across the machine. Applying the energy equation between the two sections and expressing every term as a head of water, $$H = \frac{p_1-p_2}{\rho g} + (z_1-z_2) + \frac{V_1^2-V_2^2}{2g}.$$ The pressure term is $226\,000/(1000 \times 9.81) = 23.04$ m for the inlet and the outlet gauge is already a head, $-4.5$ m, so their difference is 27.54 m. The elevation term is $+5.0$ m and the velocity term is $(11.00^2-9.20^2)/(2 \times 9.81) = 1.86$ m. Hence $$H = 27.54 + 5.00 + 1.86 = 34.40\ \text{m}.$$
Part II — hydraulic power and efficiency. $$P_{water} = \rho g Q H = 1000 \times 9.81 \times 354 \times 34.40 = 119.45\ \text{MW},$$ $$\boxed{\eta = \frac{110}{119.45} = 0.921 = 92.1\ \%}$$ This is the combined turbine-and-generator efficiency, and it sits squarely in the 90–94 % band expected of a large Kaplan unit near its design point — the sanity check that the head was assembled with the right signs.
Check: the Page 13 power curve is quoted at an air density of 1.225 kg/m³ and that value is used throughout Part I so that the computed and specified outputs are directly comparable. The constants page gives 1.21 kg/m³ at 15 °C; using it instead lowers every power in Part I by 1.2 % and leaves all four efficiencies unchanged. The two chart readings (58 % and 44 %) carry about ±1 point of graph-reading uncertainty, and 1150 kW is read to the nearest 50 kW.
Question 5 Part II — control volume across the Mactaquac unit. The useful head is the sum of the pressure, elevation and velocity heads between the two measuring sections; the electrical output is what leaves the generator terminals.
