22-Mec-B3 Energy Conversion and Power Generation · May 2014

Question 4 of 6: Wind and Water Power

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. Closed book, three hours. Two sections: Section A (Calculative) Questions 1–4 and Section B (Descriptive) Questions 5–6. Candidates answer three questions from Section A and one from Section B; four questions of 15 marks each constitute a complete 60-mark paper. Reference data for particular questions are bound in as pages 10–17, reference formulae and constants as pages 18–21, and the steam tables from Granet & Bluestein are supplied. All six printed questions are solved below.

Reference texts for 22-Mec-B3 Energy Conversion and Power Generation

Granet, I. and Bluestein, M., Thermodynamics and Heat Power, 6th ed. — the source of the steam tables (Tables A.1–A.4 SI) bound into this paper.
El-Wakil, M. M., Powerplant Technology — steam-station cycles, heat balance diagrams, gas turbines, wind and solar conversion (Figure 5.6 of this text is paper page 14).
Çengel, Y. A. and Boles, M. A., Thermodynamics: An Engineering Approach, 9th ed. — Rankine, Brayton, Otto, Diesel and combined cycles.
Fox, R. W., Pritchard, P. J. and McDonald, A. T., Introduction to Fluid Mechanics, 10th ed. — the energy equation and head-loss accounting used for the hydro station.
Rayaprolu, K., Boilers for Power and Process — boiler efficiency and heat-rate conventions.

Units and conventions used throughout. The paper's own constant sheet (page 19) is used without substitution: $g = 9.81\ \text{m/s}^2$, $c_p$ air $= 1.005\ \text{kJ/kg}\cdot \text{K}$, $c_v$ air $= 0.718\ \text{kJ/kg}\cdot\text{K}$, $c_p$ water $= 4.190\ \text{kJ/kg} \cdot\text{K}$, $\rho_{\text{water}} = 1000\ \text{kg/m}^3$, $p_{\text{atm}} = 100\ \text{kPa}$. Monetary amounts are in Canadian dollars, as the paper is an Engineers Canada national examination.

Question 4: Wind and Water Power (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.

Given.

Part	Quantity	Value
I	Rotor diameter and swept area	80 m ; 5 027 m²
I	Rotor speed of revolution	15.7 rev/min
I	Wind speed and air density (page 13 power curve)	10 m/s ; 1.225 kg/m³
I	Page-14 curve readings at the computed tip-speed ratio	ideal 58 % ; actual 43 %
I	Page-13 power curve at 10 m/s	≈ 1 150 kW
II	Elevations, points 1 / 2 / 3 / 4	1170.5 / 1091.5 / 1086.5 / 1094.7 m
II	Penstock and draft-tube diameters	7.0 m ; 5.0 m
II	Gauge pressures, points 2 and 3	700 kPa ; 65 kPa
II	Volume flow rate	200 m³/s

Find. Part I: the cascade from wind kinetic power down to the manufacturer's declared output, with the efficiency claimed at each step. Part II: the two waterway velocities, the head lost upstream and downstream of the machine, and the plant's gross and turbine hydraulic powers.

Figure 4.1 — Part I answered as a cascade. Each bar is what survives the previous loss mechanism: the Betz limit removes the kinetic energy the air must retain to leave the disc, the ideal-propeller curve adds swirl in the wake, the actual-rotor curve adds profile and tip-vortex drag, and the manufacturer's figure adds gearbox and generator losses.

Approach. Part I is a chain of multiplications on the same base quantity $\tfrac{1}{2}\rho A V^3$, so the base is computed once and each part supplies a coefficient; the only real work is getting the tip-speed ratio right so the graph is read at the correct abscissa. Part II is one application of the steady-flow energy equation in head form between four stations, with head loss taken as the deficit in total head.

Part I (a) — kinetic energy and available power. Specific kinetic energy is a property of the wind alone: $$\text{ke} = \tfrac{1}{2}V^2 = \tfrac{1}{2}(10)^2 = \boxed{50\ \text{J/kg}}$$ The mass flow through the swept area (a check first: $\pi D^2/4 = \pi (80)^2/4 = 5\,027$ m², confirming the datasheet figure) is $$\dot m = \rho A V = 1.225 \times 5\,027 \times 10 = 61\,580\ \text{kg/s}$$ so the potential power in that stream is $$P_{\text{avail}} = \dot m \cdot \text{ke} = \tfrac{1}{2}\rho A V^3 = \tfrac{1}{2} \times 1.225 \times 5\,027 \times 10^3 = \boxed{3.079\ \text{MW}}$$ The cube law is the dominant fact of wind engineering: doubling the wind speed multiplies the resource eightfold, which is why siting matters more than machine refinement.
Part I (b) — maximum theoretical power and efficiency. The paper's formula sheet (page 20) gives the actuator-disc result directly: $$P_{\max} = \frac{8 \rho A V_1^3}{27} = \frac{8 \times 1.225 \times 5\,027 \times 10^3}{27} = \boxed{1.825\ \text{MW}}$$ $$\eta_{\max} = \frac{P_{\max}}{P_{\text{avail}}} = \frac{8/27}{1/2} = \frac{16}{27} = \boxed{0.593\ (59.3\ \%)}$$ This is the Betz limit, and it holds for any wind speed and any rotor because the derivation uses only mass, momentum and energy conservation across a disc. Its physical content is that the air must still be moving when it leaves — extracting all the kinetic energy would stop the flow and therefore stop the supply — and the optimum compromise leaves the wake at one third of the free stream velocity.
Part I (c) — ideal efficiency and power at the actual tip-speed ratio. The rotor turns at 15.7 rev/min, so the blade tip travels at $$u_{\text{tip}} = \frac{\pi D N}{60} = \frac{\pi \times 80 \times 15.7}{60} = 65.8\ \text{m/s} \qquad \lambda = \frac{u_{\text{tip}}}{V} = \frac{65.8}{10} = 6.58$$ Entering page 14 at $\lambda = 6.58$ on the "ideal efficiency for propeller-type windmills" curve gives $$\eta_{\text{ideal}} \approx 58\ \% \;\Rightarrow\; P_{\text{ideal}} = 0.58 \times 3.079 = \boxed{1.79\ \text{MW}}$$ This curve sits just below the Betz value because it charges the rotor for the rotational kinetic energy left in the wake; that penalty shrinks as $\lambda$ rises, which is exactly why the curve climbs steeply to $\lambda \approx 3$ and then flattens towards 59 %.
Part I (d) — actual efficiency and power at the same tip-speed ratio. Reading the real-machine band of page 14 at $\lambda = 6.58$, $$\eta_{\text{actual}} \approx 43\ \% \;\Rightarrow\; P_{\text{actual}} = 0.43 \times 3.079 = \boxed{1.32\ \text{MW}}$$ The 15-point gap from the ideal curve is aerofoil profile drag and tip-vortex loss, neither of which the momentum theory sees.
Part I (e) — the manufacturer's figure, and the comparison the question asks for. Scaling the page-13 power curve at 10 m/s gives $$P_{\text{spec}} \approx \boxed{1\,150\ \text{kW} = 1.15\ \text{MW}}$$ which is 37.3 % of the 3.079 MW in the wind. Set against the four preceding numbers the cascade reads $3.079 \rightarrow 1.825 \rightarrow 1.79 \rightarrow 1.32 \rightarrow 1.15$ MW, and the last step is the 87 % combined efficiency of the gearbox, the asynchronous generator and the switchgear — entirely credible for a geared machine with a slip generator. Two further observations close the comparison. The V80 is rated 1.8 MW but reaches that only at its 16 m/s nominal wind speed, so 1.15 MW at 10 m/s is a normal part-load point, not a shortfall. And the datasheet's "nominal output 1.0 MW" line refers to the alternative 1.0 MW generator offered on the same platform, not to this machine.

Check: reading page 14 above the plotted range of the three-blade curve. The "modern three-blade type" curve on page 14 is drawn only to $\lambda \approx 5.6$, where it reads 41 % after peaking near 43.5 % at $\lambda \approx 4.3$. The V80's operating point at $\lambda = 6.58$ therefore lies beyond it, and the actual reading of 43 % taken above is from the one real-machine curve that is plotted there (the high-speed two-blade type, peaking at 46.5 % near $\lambda = 5.9$). Extrapolating the three-blade curve instead would give roughly 37 %, which is precisely the 37.3 % that the manufacturer's own power curve implies — a satisfying consistency, and the reason the answer to (e) sits below the answer to (d) by more than gearbox losses alone would explain. Both readings are defensible on the chart supplied; the 43 % figure is boxed because the question asks what the graph shows at the stated ratio.

Figure 4.2 — Part II. The dashed line is the energy grade line: it starts at the reservoir surface, drops by the penstock loss to station 2, falls steeply across the turbine, and drops again by the draft-tube and tailrace loss to the tailwater surface. Note that station 3 lies below tailwater level, so the draft tube runs under positive gauge pressure.

Part II (a)(i) — velocities at the turbine inlet and outlet. From continuity at the two stated diameters, $$A_2 = \frac{\pi (7.0)^2}{4} = 38.49\ \text{m}^2 \qquad V_2 = \frac{Q}{A_2} = \frac{200}{38.49} = \boxed{5.20\ \text{m/s}}$$ $$A_3 = \frac{\pi (5.0)^2}{4} = 19.63\ \text{m}^2 \qquad V_3 = \frac{Q}{A_3} = \frac{200}{19.63} = \boxed{10.19\ \text{m/s}}$$ Five metres per second in a penstock is conventional practice; much above that and the friction loss, which grows as $V^2$, starts to eat the head the dam was built to provide.
Part II (a)(ii) — head loss in the penstock, stations 1 to 2. Total head at each station is elevation plus pressure head plus velocity head. At the reservoir surface the pressure is atmospheric and the velocity negligible, so $$H_1 = z_1 = 1170.50\ \text{m}$$ At the turbine inlet, with 700 kPa gauge and 5.20 m/s, $$H_2 = z_2 + \frac{p_2}{\rho g} + \frac{V_2^2}{2g} = 1091.50 + \frac{700\,000}{1000 \times 9.81} + \frac{(5.197)^2}{2 \times 9.81}$$ $$H_2 = 1091.50 + 71.36 + 1.38 = 1164.23\ \text{m}$$ The deficit in total head is the loss: $$h_{L,1\text{-}2} = H_1 - H_2 = 1170.50 - 1164.23 = \boxed{6.27\ \text{m}}$$ That is 8.3 % of the gross head, consumed in the intake gate, the bend at the flexible-joint chamber and the friction of the steel-lined penstock.
Part II (a)(ii) continued — head loss in the draft tube and tailrace, stations 3 to 4. At the draft-tube inlet, with 65 kPa gauge and 10.19 m/s, $$H_3 = z_3 + \frac{p_3}{\rho g} + \frac{V_3^2}{2g} = 1086.50 + 6.63 + 5.29 = 1098.41\ \text{m}$$ and at the tailwater surface $H_4 = z_4 = 1094.70$ m, so $$h_{L,3\text{-}4} = H_3 - H_4 = 1098.41 - 1094.70 = \boxed{3.71\ \text{m}}$$ Most of this is the kinetic energy still in the water at station 3: the draft tube's job is to convert that 5.29 m of velocity head back into pressure before discharge, and it recovers only part of it. A sanity check worth quoting is that station 3 sits 8.2 m below tailwater level, and the 6.63 m of gauge pressure there is the hydrostatic 8.2 m less the 1.6 m already converted to velocity — consistent, and confirming the sign convention.
Part II (b)(i) — potential power of the whole plant. On elevation difference and flow alone, $$H_{\text{gross}} = z_1 - z_4 = 1170.50 - 1094.70 = 75.80\ \text{m}$$ $$P_{\text{pot}} = \rho g Q H_{\text{gross}} = 1000 \times 9.81 \times 200 \times 75.80 = 148.7 \times 10^{6}\ \text{W} = \boxed{148.7\ \text{MW}}$$ This is the resource the site offers, before any of it is lost in the waterways or the machine.
Part II (b)(ii) — hydraulic power developed in the turbine. Between stations 2 and 3 the head available to the runner is the difference in total head, $$H_{\text{turb}} = H_2 - H_3 = 1164.23 - 1098.41 = 65.82\ \text{m}$$ $$P_{\text{turb}} = \rho g Q H_{\text{turb}} = 1000 \times 9.81 \times 200 \times 65.82 = \boxed{129.1\ \text{MW}}$$ The arithmetic closes exactly, which is the check to quote: the two head losses account for the whole difference, $$h_{L,1\text{-}2} + h_{L,3\text{-}4} = 6.27 + 3.71 = 9.98\ \text{m} = H_{\text{gross}} - H_{\text{turb}} = 75.80 - 65.82$$ so the waterway efficiency is $65.82/75.80 = 86.8\ \%$. Multiplying by a runner and generator efficiency of about 92 % would give roughly 119 MW at the terminals, which is the order of the real Van der Kloof station's four units.

Part	Quantity	Result
I (a)	Specific kinetic energy ; available power	50 J/kg ; 3.079 MW ($\dot m = 61\,580$ kg/s)
I (b)	Betz maximum power ; efficiency	1.825 MW ; 59.3 % (16/27)
I (c)	Tip-speed ratio ; ideal efficiency and power	$\lambda = 6.58$ ; 58 % ; 1.79 MW
I (d)	Actual efficiency and power	43 % ; 1.32 MW
I (e)	Manufacturer's output at 10 m/s	1.15 MW (37.3 % overall)
II (a)(i)	Velocities at points 2 and 3	5.20 m/s ; 10.19 m/s
II (a)(ii)	Head loss 1–2 ; 3–4	6.27 m ; 3.71 m
II (b)(i)	Potential power on gross head 75.80 m	148.7 MW
II (b)(ii)	Hydraulic power on 65.82 m across the turbine	129.1 MW (waterway efficiency 86.8 %)

Topic. The actuator-disc (Betz) limit and the efficiency cascade of a real wind turbine; the energy equation in head form applied to a hydro waterway.

Key relations. Wind: $P_{\text{avail}} = \tfrac{1}{2}\rho A V^3$; $P_{\max} = \tfrac{8}{27}\rho A V^3$ so $\eta_{\max} = 16/27$; $\lambda = \pi D N / (60 V)$; $P = \eta(\lambda) \cdot \tfrac{1}{2}\rho A V^3$. Hydro: $H = z + p/(\rho g) + V^2/(2g)$; $h_L = H_{\text{up}} - H_{\text{down}}$; $P = \rho g Q H$; $V = Q/A$.

Why this works. The Betz analysis puts a control volume around the rotor disc and writes mass, momentum and energy conservation for the streamtube. Thrust from the momentum equation and power from the energy equation both depend on the induction factor, and maximising their product gives an axial induction of one third and a power coefficient of 16/27. Everything the momentum theory cannot see — wake swirl, aerofoil drag, tip vortices, gearbox and generator loss — then appears as successive multipliers on the same $\tfrac{1}{2}\rho A V^3$, which is why the answer is naturally written as a cascade. On the hydro side, total head is mechanical energy per unit weight, so a drop in total head between two stations with no machine between them can only be irreversibility; inserting the machine, the drop is the work extracted. The two accountings must add to the gross head, and checking that they do validates every reading taken off the drawing.

Common pitfall. In Part I, reading the graph at the wrong abscissa — the tip-speed ratio must be built from the rotor's own rev/min and diameter, not guessed — and confusing the Betz limit with a machine efficiency. In Part II, the classic error is computing turbine power on the elevation difference $z_2 - z_3$ (only 5 m) instead of on the total-head difference, which omits the pressure and velocity heads that carry almost all the energy at those two stations. Forgetting that the stated pressures are gauge, or omitting the velocity head at station 3 where it is 5.3 m, are the other two.

Source. El-Wakil, Powerplant Technology, Ch. 5 (hydroelectric plant) and Ch. 15 (wind energy; Figure 5.6 reproduced as paper page 14, after Wilson and Lissaman); Fox, Pritchard & McDonald, Introduction to Fluid Mechanics, 10th ed., §8-8 and Ch. 10 (turbomachinery, head accounting).

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