Hydro Turbine Model (Vanderkloof): 22-Mec-A6 Fluid Machinery, May 2015, Q2

Question 2: Hydro Turbine Model (Vanderkloof) (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 runner data for a large Francis machine and its scale test rig.

Given data
Prototype output / speed	120 MW / 125 rev/min
Net head / flow	65 m / 200 m³/s
Prototype runner diameter	5.462 m
Model runner diameter / head	0.200 m / 10 m
Prototype electrical efficiency	98 %

Find. Specific speed, overall efficiency, model speed/flow/power, and the model hydraulic efficiency target.

Approach. Use the dimensionless power specific speed from the reference sheet, the affinity (similarity) laws to scale speed, flow and power to the model, then the full Moody equation to relate the two hydraulic efficiencies at different heads.

Specific speed. $\Omega_{sp}=\dfrac{\omega\,P^{1/2}}{\rho^{1/2}(gH)^{5/4}}$ with $\omega=2\pi(125)/60=13.09\ \text{rad/s}$ gives $\boxed{\Omega_{sp}=1.42}$ — a Francis machine.
Overall efficiency. $\eta_o=\dfrac{P}{\rho g Q H}=\dfrac{120\times10^{6}}{1000\cdot9.81\cdot200\cdot65}=0.941$.
Model speed. Equal head and flow coefficients give $N_m=N\dfrac{D_p}{D_m}\sqrt{\dfrac{H_m}{H_p}}=125\cdot\dfrac{5.462}{0.200}\sqrt{\dfrac{10}{65}}=1339\ \text{rev/min}$.
Model flow. $Q_m=Q\dfrac{N_m}{N}\left(\dfrac{D_m}{D_p}\right)^{3}=0.105\ \text{m}^3/\text{s}$.
Ideal model power. $P_m=\rho g Q_m H_m=1000\cdot9.81\cdot0.105\cdot10=10.3\ \text{kW}$.
Required model efficiency. The prototype hydraulic efficiency needed is $\eta_{P}=\eta_o/0.98=0.960$. The full Moody relation $\eta_P=1-(1-\eta_M)\left(\tfrac{D_m}{D_p}\right)^{1/4}\left(\tfrac{H_m}{H_p}\right)^{1/10}$ inverts to $\boxed{\eta_M=0.890}$.

The affinity laws hold because a homologous model runs at the same non-dimensional operating point; the Moody equation then corrects for the fact that a small model has proportionally larger boundary layers and therefore lower hydraulic efficiency. Because the model head (10 m) differs from the prototype (65 m), the full form is used; the approximate diameter-only form $\eta_M\approx1-(1-\eta_P)(D_p/D_m)^{1/5}$ gives about 92.3 % and is quoted for comparison only.

Question 2 results
Quantity	Value
(a) Specific speed Ω_sp	1.42 (Francis)
(b) Overall efficiency	94.1 %
(c) Model speed	1339 rev/min
(d) Model flow	0.105 m³/s
(e) Ideal model power	10.3 kW
(f) Required model efficiency	89.0 %

Topic: Turbine similarity (affinity) laws and the Moody scale-up

Key relations: Head coeff $gH/N^{2}D^{2}$, flow coeff $Q/ND^{3}$, power coeff $P/\rho N^{3}D^{5}$; Moody $\eta_P=1-(1-\eta_M)(D_m/D_p)^{1/4}(H_m/H_p)^{1/10}$.

Why this works: Geometric similarity plus equal Reynolds regime lets a model at a different size and head reproduce the prototype flow field, so the dimensionless coefficients match and speed, flow and power scale by fixed powers of the diameter and head ratios. Efficiency does not scale identically because viscous losses are relatively larger on the small model — the Moody equation supplies the correction.

Common pitfall: Using the diameter-only Moody form when the model head differs from the prototype, which understates the required model efficiency.

Source: Dixon & Hall, 7th ed., §2 (dimensional analysis) and Ch. 9; Moody scale relation.

Question 2 of 8: Hydro Turbine Model (Vanderkloof)

Question 2: Hydro Turbine Model (Vanderkloof) (10 marks)