Question 2 of 8: Hydro Turbine Model

Question 2: Hydro Turbine Model (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. Prototype and model:

The machine is homologous — geometrically identical at 1:23.28 scale. The remaining specifications on the page (5225 MW plant capacity, eleven units, 15 kV, 60 Hz, 77.13 Mg runner) are context, not data for the calculation; the 60 Hz and 200 rev/min are consistent with a 36-pole generator, which is the only cross-check they offer.

Find. The specific speed and the chart efficiency of the prototype, the flow it must pass, then the model's speed, flow, ideal power, and the efficiency the model test would have to measure for the full-size machine to reach its guarantee.

Approach. Specific speed from page 21 to enter the page-13 efficiency chart, then P = ηρgQH for the flow; the two similarity equations of page 21 for the model speed and flow, ρgQH again for its ideal power, and finally the Moody relation inverted for the model efficiency.

1. Part (a) — power specific speed. This is the same machine as Question 1 Part II, so the same evaluation applies: $$N_s=\frac{\omega\sqrt{P}}{\rho^{1/2}(gH)^{5/4}} =\frac{20.94\sqrt{483.2\times 10^{6}}}{\sqrt{1000}\,(9.81\times 312.4)^{5/4}} =\boxed{0.638\ \text{rad}}$$ Getting the same number twice from two differently worded questions is itself a check that the formula has been entered correctly.
2. Part (b) — chart efficiency. Entering the page-13 plot of η_max against N_St at 0.638 puts the machine in the middle of the Francis band, close to the crest of the Francis curve: $$\boxed{\eta \approx 0.93}$$ That is where a large Francis unit belongs; the highest-efficiency hydraulic machines ever built are Francis runners at roughly this specific speed, and 93 per cent for the runner alone is realistic.
3. Part (c) — prototype volume flow. From P = ηρgQH (page 21, with the efficiency inserted because 483.2 MW is shaft output, not water power): $$Q=\frac{P}{\eta\rho gH}=\frac{483.2\times 10^{6}}{0.93\times 1000\times 9.81\times 312.4} =\boxed{169.5\ \text{m}^3\text{/s}}$$ Eleven such units pass about 1865 m³/s between them, which is the right order for the Churchill River.
4. Part (c, second) — the speed the model must run at. Homologous machines run at the same head coefficient, which is the second similarity equation of page 21: $$\frac{H_M}{H_P}=\left(\frac{\omega_M}{\omega_P}\right)^{2}\left(\frac{D_M}{D_P}\right)^{2} \;\Longrightarrow\; \frac{\omega_M}{\omega_P}=\sqrt{\frac{H_M}{H_P}}\;\frac{D_P}{D_M}$$ Substituting the numbers, $$\frac{\omega_M}{\omega_P}=\sqrt{\frac{20.0}{312.4}}\times\frac{5.82}{0.250} =0.2530\times 23.28=5.890$$ so $$\boxed{N_M=200\times 5.890=1178\ \text{rev/min}}$$ The model runs almost six times faster than the prototype. That is the usual outcome and it is what makes model testing practical: a small runner on a modest head still reaches full-size velocity triangles.
5. Part (d) — the flow the model rig must supply. The first similarity equation of page 21 gives the flow coefficient: $$Q_M=Q_P\left(\frac{\omega_M}{\omega_P}\right)\left(\frac{D_M}{D_P}\right)^{3} =169.5\times 5.890\times\left(\frac{0.250}{5.82}\right)^{3}$$ With (0.250/5.82)³ = 7.926 × 10⁻⁵, $$\boxed{Q_M=0.0792\ \text{m}^3\text{/s}=79.2\ \text{L/s}}$$ Eighty litres per second is an entirely ordinary laboratory duty, which is the point of the exercise.
6. Part (e) — ideal power of the model. “Ideal, no friction” means the full water power is counted, so no efficiency appears: $$P_M=\rho gQ_MH_M=1000\times 9.81\times 0.0792\times 20.0=\boxed{15.5\ \text{kW}}$$ As a cross-check the power affinity law of page 21, P_M/P_P = (ω_M/ω_P)³(D_M/D_P)⁵, applied to the prototype's water power of 519.6 MW returns the same 15.5 kW.
7. Part (f) — the efficiency the model must demonstrate. Model and prototype do not share an efficiency because the Reynolds numbers and the relative roughness differ; the small machine is proportionally rougher and proportionally more viscous, so it always performs worse. The Moody relation on page 21 quantifies this: $$\eta_P=1-(1-\eta_M)\left(\frac{D_M}{D_P}\right)^{1/4}\left(\frac{H_M}{H_P}\right)^{1/10}$$ Solving for the model efficiency, $$1-\eta_M=\frac{1-\eta_P}{(D_M/D_P)^{1/4}(H_M/H_P)^{1/10}}$$ The scaling factor is $$(0.042955)^{1/4}(0.064020)^{1/10}=0.4553\times 0.7597=0.3458$$ so $$1-\eta_M=\frac{1-0.93}{0.3458}=0.2024 \qquad\Longrightarrow\qquad \boxed{\eta_M=0.798\ \ (79.8\ \text{per cent})}$$ The result reads oddly at first and is the whole point of the sub-part: a model that measures only 79.8 per cent is nevertheless proof that the full-size machine will reach 93 per cent. Ask the model to hit 93 per cent itself and no runner on earth would pass the test.