Question 4 of 8: Wind and water power

Question 4: Wind and water power (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, from the page-12 data sheet: rotor diameter 80 m, swept area 5027 m², rotor speed 15.7 rev/min, three blades, air density 1.225 kg/m³, and a power curve reading about 1150 kW at 10 m/s. Part II, the seven measurements listed in the question.

Find. Part I: the Betz power and efficiency, the ideal and actual chart efficiencies and powers at the machine's own tip-speed ratio, and the manufacturer's stated output, all at 10 m/s. Part II: the efficiency of the Mactaquac turbine-generator set.

Approach. Part I is a chain of efficiencies applied to one number, the power flux through the swept area; the only real work is computing the tip-speed ratio so the chart can be entered. Part II is a single application of the energy equation between the two pressure tappings, with pressure, elevation and velocity heads all counted, followed by a comparison of water power with electrical output.

1. Part I (a) — the power in the wind and the Betz limit. The kinetic energy flux through the swept area is $$P_{\text{wind}} = \tfrac{1}{2}\rho A V_1^{3} = 0.5 \times 1.225 \times 5027 \times 10^{3} = 3\,079\ \text{kW}$$ and the reference sheet gives the maximum ideal power a rotor can extract as $P_{\max} = 8\rho A V_1^{3}/27$: $$P_{\max} = \frac{8 \times 1.225 \times 5027 \times 10^{3}}{27} = \boxed{1825\ \text{kW}}, \qquad \eta_{\max} = \frac{P_{\max}}{P_{\text{wind}}} = \frac{16}{27} = \boxed{59.3\ \%}$$ This is the Betz limit. It comes from energy and momentum together: the rotor can only extract power by slowing the air, but if it slows it too much the mass flow through the disc collapses, and the optimum is an exit velocity of one third of the free-stream value.
2. Part I (b) — enter the chart at the machine's own tip-speed ratio. The blade tip travels one circumference per revolution: $$u = \pi D N/60 = \pi \times 80 \times 15.7/60 = 65.76\ \text{m/s}, \qquad \lambda = \frac{u}{V_1} = \frac{65.76}{10} = 6.58$$ Reading the ideal propeller curve on page 13 at $\lambda = 6.58$ gives about 58 %, so $$\eta_{\text{ideal}} = \boxed{58\ \%}, \qquad P_{\text{ideal}} = 0.58 \times 3079 = \boxed{1786\ \text{kW}}$$ The ideal curve is still climbing gently toward the Betz value at this λ, because at high tip-speed ratio the wake swirl that the ideal curve accounts for becomes negligible.
3. Part I (c) — the actual (real-machine) reading. At $\lambda = 6.58$ the V80 is running past the plotted end of the modern three-blade curve, which finishes near $\lambda = 5.6$. The real-machine band at 6.58 is given by the high-speed two-blade curve at about 43 %: $$\eta_{\text{actual}} = \boxed{43\ \%}, \qquad P_{\text{actual}} = 0.43 \times 3079 = \boxed{1324\ \text{kW}}$$ Extrapolating the three-blade curve instead would give roughly 37 %, or 1139 kW; both readings are quoted here because the chart genuinely does not cover this λ for a three-blade rotor.
4. Part I (d) — what the manufacturer actually claims. The page-12 power curve, drawn for the same 1.225 kg/m³ air density, reads $$P_{\text{spec}} = \boxed{1150\ \text{kW}\ \text{at}\ 10\ \text{m/s}}$$ which is an overall efficiency of $1150/3079 = 37.4\ \%$. Comparing the four numbers: the Betz limit allows 1825 kW, the ideal chart curve 1786 kW, the real-rotor band 1324 kW and the machine delivers 1150 kW. The step from 1324 kW to 1150 kW is the gearbox, generator and converter train, about 87 % combined — and the fact that the manufacturer's 37.4 % agrees with the extrapolated three-blade reading of 37 % is a strong hint that the three-blade curve, not the two-blade one, describes this rotor.
5. Part II — the velocities at the two measuring sections. Continuity gives each velocity from the common volume flow: $$A_1 = \frac{\pi (6.4)^2}{4} = 32.17\ \text{m}^2, \qquad V_1 = \frac{354}{32.17} = 11.00\ \text{m/s}$$ $$A_2 = \frac{\pi (7.0)^2}{4} = 38.48\ \text{m}^2, \qquad V_2 = \frac{354}{38.48} = 9.20\ \text{m/s}$$ The draft tube is deliberately made wider than the inlet so that some of the kinetic energy leaving the runner is recovered as pressure rather than thrown away in the tailrace.
6. Part II — the net head across the machine. The energy equation between tappings 1 and 2 gives the head available to the runner as the sum of the pressure, elevation and velocity differences: $$H = \frac{p_1 - p_2}{\rho g} + (z_1 - z_2) + \frac{V_1^2 - V_2^2}{2g}$$ The inlet pressure head is $226\,000/(1000 \times 9.81) = 23.04$ m and the outlet tapping reads −4.5 m of water, i.e. it is below atmosphere, so the pressure term is $23.04 - (-4.5) = 27.54$ m. The velocity term is $(11.00^2 - 9.20^2)/(2 \times 9.81) = 1.86$ m. Adding the 5.0 m of elevation, $$H = 27.54 + 5.00 + 1.86 = \boxed{34.40\ \text{m}}$$ The sub-atmospheric reading at the draft-tube tapping is not an error: it is the suction the draft tube is there to create, and dropping it would understate the head by 13 %.
7. Part II — water power and efficiency. The hydraulic power delivered to the machine is $$P_{\text{water}} = \rho g Q H = 1000 \times 9.81 \times 354 \times 34.40 = 119.45\ \text{MW}$$ so the efficiency measured at the generator terminals is $$\eta = \frac{P_e}{P_{\text{water}}} = \frac{110}{119.45} = \boxed{0.921 = 92.1\ \%}$$ This is the combined turbine-and-generator efficiency, because the measured output is electrical. If the generator alone is taken as 98.5 % efficient, the runner itself is working at 93.5 %, which is squarely in the expected band for a large Kaplan machine.
8. Part II — confirm the machine type from its specific speed. $$N_s = \frac{N\sqrt{P}}{H^{1.25}} = \frac{112.5\sqrt{110\,000}}{34.40^{1.25}} = 448$$ (metric power form, P in kW, H in m). Values above about 300 belong to axial-flow machines, so a Kaplan runner is indeed the right choice for this head and flow — a Francis runner at 34 m would have to turn much more slowly.