Question 3 of 8: Hydro Turbines

Question 3 — Hydro Turbines (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.

Part I — Kaplan turbine efficiency

Given.

Find. The pipe velocities at the two measuring stations, the hydraulic power the water delivers to the machine, and the overall turbine-generator efficiency.

Approach. Take a control volume between the two pressure tappings, evaluate the total head extracted from the energy equation of page 20 (pressure head plus elevation plus velocity head), convert it to power with $P = \rho g Q H$, and divide the measured electrical output by it.

1. Flow velocities at the two measuring stations (part a). Continuity through circular sections gives $$A_1 = \frac{\pi}{4}(6.4)^2 = 32.17\ \text{m}^2,\qquad A_2 = \frac{\pi}{4}(7.0)^2 = 38.48\ \text{m}^2$$ $$\boxed{V_1 = \frac{354}{32.17} = 11.00\ \text{m/s}\qquad V_2 = \frac{354}{38.48} = 9.20\ \text{m/s}}$$ The draft tube is doing its job: it has already recovered part of the runner exit kinetic energy by the time the outlet tapping is reached.
2. Head extracted between the tappings (part b). The energy equation of page 20, written between stations 1 and 2 with the work term collected on one side, makes the head delivered to the machine $$H = \frac{p_1 - p_2}{\rho g} + (z_1 - z_2) + \frac{V_1^2 - V_2^2}{2g}$$ Taking each term in turn: the pressure head is $\dfrac{226\,000}{1000 \times 9.81} - (-4.5) = 23.04 + 4.50 = 27.54\ \text{m}$; the elevation term is $+5.00\ \text{m}$ because the outlet tapping is lower; and the velocity-head term is $$\frac{11.004^2 - 9.199^2}{2(9.81)} = \frac{121.09 - 84.61}{19.62} = 1.86\ \text{m}$$ Adding them, $H = 27.54 + 5.00 + 1.86 = 34.40\ \text{m}$. The negative outlet pressure is not an error to be discarded — it is the sub-atmospheric pressure at the draft-tube tapping, and it adds to the head the machine can use.
3. Hydraulic power input (part b concluded). The page-21 relation converts head to power: $$P_{\text{hyd}} = \rho\,g\,Q\,H = 1000 \times 9.81 \times 354 \times 34.397$$ $$\boxed{P_{\text{hyd}} = 119.5\ \text{MW}}$$
4. Electrical output and overall efficiency (part c). The electrical output is the measured value, $P_{el} = 110\ \text{MW}$, so the efficiency of the turbine-generator set taken together is $$\eta = \frac{P_{el}}{P_{\text{hyd}}} = \frac{110}{119.45}$$ $$\boxed{\eta = 0.921 = 92.1\ \text{per cent}}$$ That figure is exactly where a large Kaplan unit with its generator should sit at full gate, which is the real point of the exercise: the measurement set is a plausible acceptance test, and a result of 92 per cent tells the operator the machine is healthy. A result of 75 per cent would point to a bad tapping or a fouled runner, not to a bad turbine design.

Part II — 4 MW hydro plant

Given.

Find. The flow at full power on the low-reservoir condition, the corresponding penstock velocity, whether the velocity rises or falls when the reservoir fills, and its value in that condition.

[Figure not reproduced: The plant section redrawn from the page-14 attachment with the levels that matter. The gross head at the design condition is the 15.67 m between normal low reservoir level and full-load tailrace level; the penstock diameter is scaled against the drawing's 18.52 m turbine-setting dimension. See the official exam paper.]

Approach. With friction neglected the gross head between the two free surfaces is the head available at the machine, so the flow follows from $P = \eta\rho g Q H$ once an efficiency has been chosen; continuity through the scaled penstock area then gives the velocity, and the same argument repeated at the higher reservoir level answers parts (c) and (d).

1. Choose a turbine efficiency and take the design head. The question asks for an appropriate value. A 4 MW Kaplan unit in a low-head application, complete with its generator, will run at about 90 per cent at full gate — consistent with the 92.1 per cent found for the much larger Mactaquac machine in Part I, since efficiency falls slowly with size (the Moody relation of page 21 makes the same point). Take $\eta = 0.90$. The gross head between normal low reservoir level and full-load tailrace level is $$H_a = 42.97 - 27.30 = 15.67\ \text{m}$$
2. Flow rate at maximum power (part a). Rearranging $P = \eta\,\rho\,g\,Q\,H$, $$Q_a = \frac{P}{\eta\,\rho\,g\,H_a} = \frac{4 \times 10^{6}}{0.90 \times 1000 \times 9.81 \times 15.67}$$ $$\boxed{Q_a = 28.9\ \text{m}^3/\text{s}}$$ Twenty-nine cubic metres a second for four megawatts is the signature of a low-head machine, and it is why the Kaplan — a high specific-speed, high-flow runner — is the right choice for this site.
3. Scale the penstock diameter from the drawing. The only dimension printed on the page-14 section is the 18.52 m turbine setting, so it is the scale. Measuring the penstock across its clear bore against that dimension gives a ratio of about $0.151$, hence $$D_p \approx 0.151 \times 18.52 = 2.8\ \text{m},\qquad A_p = \frac{\pi}{4}(2.8)^2 = 6.16\ \text{m}^2$$
4. Penstock velocity at the design condition (part b). Continuity gives $$\boxed{V_a = \frac{Q_a}{A_p} = \frac{28.91}{6.158} = 4.70\ \text{m/s}}$$ This sits in the usual 3–6 m/s band for a short steel penstock, where the cost of steel and the cost of friction head are roughly balanced — a useful confirmation that the scaled diameter is sensible.
5. Effect of a higher reservoir level, with reasons (part c). With the tailrace unchanged at 27.30 m, filling the reservoir to normal high water level raises the gross head to $$H_c = 45.54 - 27.30 = 18.24\ \text{m}$$ The power demanded of the machine is still 4 MW, and power is the product of head and flow. Since the head has risen by about 16 per cent and the power is fixed, the flow — and therefore the velocity in the penstock, whose area is fixed — must fall by very nearly the same proportion. Physically the wicket gates and runner blades of the Kaplan close down to admit less water because each cubic metre now carries more energy. The velocity in the penstock will therefore be less than the 4.70 m/s of part (b).
6. Penstock velocity at the high-reservoir condition (part d). Repeating the calculation with the new head, $$Q_c = \frac{4 \times 10^{6}}{0.90 \times 1000 \times 9.81 \times 18.24} = 24.84\ \text{m}^3/\text{s}$$ $$\boxed{V_c = \frac{24.84}{6.158} = 4.03\ \text{m/s}}$$ The velocity has fallen by 14 per cent, as the reasoning in part (c) predicted, and the friction loss in the penstock (which varies as the square of velocity) falls by a quarter, so neglecting it is even safer at this condition than at the design one.