Hydro Turbine Model: 22-Mec-A6 Fluid Machinery, December 2013, Q2

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. The prototype and its homologous model share the design and test data tabulated below.

Given data (prototype / model)
Generator output	120 MW	Net head H_P	65 m
Speed N_P	125 rev/min	Design flow Q_P	200 m³/s
Runner dia D_P	5.462 m	Model dia D_M	0.200 m
Model head H_M	10 m	Prototype elec. eff.	98%

Find. Dimensionless specific speed and turbine type; overall efficiency; homologous model speed, flow and ideal power; and the model hydraulic efficiency target via the Moody scale-up.

Approach. Use the reference-sheet dimensionless power specific speed to classify the machine; overall efficiency from output over water power; the similarity groups $gH/N^2D^2$ and $Q/ND^3$ for the model conditions; and the full Moody equation to relate model and prototype hydraulic efficiency.

(a) Specific speed. With $\omega=2\pi N/60=13.09\ \text{rad/s}$, the dimensionless power specific speed is $$\Omega_{sp}=\frac{\omega P^{1/2}}{\rho^{1/2}(gH)^{5/4}}=\frac{13.09\,(120\times10^6)^{1/2}}{1000^{1/2}(9.81\cdot 65)^{5/4}}=\boxed{1.42}$$A value near 1.4 (rad) lies in the Francis range — consistent with the 65 m head.
(b) Overall efficiency. $$\eta_{ov}=\frac{P_{elec}}{\rho g Q_P H_P}=\frac{120\times10^{6}}{1000\cdot 9.81\cdot 200\cdot 65}=\boxed{94.1\%}$$
(c) Model speed. Equal head coefficient $gH/(N^2D^2)$ gives $$N_M=N_P\frac{D_P}{D_M}\sqrt{\frac{H_M}{H_P}}=125\cdot\frac{5.462}{0.200}\sqrt{\frac{10}{65}}=\boxed{1339\ \text{rev/min}}$$
(d) Model flow. Equal flow coefficient $Q/(ND^3)$: $$Q_M=Q_P\frac{N_M}{N_P}\Big(\frac{D_M}{D_P}\Big)^3=200\cdot 10.71\cdot(0.03662)^3=\boxed{0.105\ \text{m}^3/\text{s}}$$
(e) Ideal model power. $$P_M=\rho g Q_M H_M=1000\cdot 9.81\cdot 0.105\cdot 10=\boxed{10.3\ \text{kW}}$$
(f) Model efficiency via Moody. The prototype must reach hydraulic efficiency $\eta_P=\eta_{ov}/\eta_{elec}=0.941/0.98=0.960$. The Moody relationship $\eta_P=1-(1-\eta_M)\left(\tfrac{D_M}{D_P}\right)^{1/4}\left(\tfrac{H_M}{H_P}\right)^{1/10}$ inverts to $$1-\eta_M=\frac{1-\eta_P}{(D_M/D_P)^{1/4}(H_M/H_P)^{1/10}}=\frac{0.0398}{0.363}\Rightarrow \eta_M=\boxed{89.0\%}$$

Check: The exam sheet also lists an approximate Moody form using the diameter ratio only, $(1-\eta_M)/(1-\eta_P)=(D_P/D_M)^{1/5}$, which gives η_M ≈ 92.3%. The full form is used above because the model head (10 m) differs from the prototype (65 m); both are acceptable exam answers if stated.

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

Topic: Turbomachine similarity and Moody scale-up

Key relations: Head coeff. $gH/N^2D^2$, flow coeff. $Q/ND^3$, power coeff. $P/\rho N^3D^5$; specific speed $\Omega_{sp}=\omega P^{1/2}/[\rho^{1/2}(gH)^{5/4}]$; Moody $\eta_P=1-(1-\eta_M)(D_M/D_P)^{1/4}(H_M/H_P)^{1/10}$.

Why this works: Geometrically similar (homologous) machines share dimensionless coefficients, so a small model at a convenient head reproduces the prototype flow field. Specific speed collapses head, flow and speed into one shape number that fixes the runner type. Efficiency is not similar — friction is relatively larger on the small model — so the Moody law scales it up with size.

Common pitfall: Cubing/fifth-powering the diameter ratio incorrectly, or assuming the model and prototype have equal efficiency, are the usual errors; also, specific speed must use a consistent (dimensionless or one national) definition.

Source: Dixon & Hall §2 (dimensional analysis, specific speed); Turton, Principles of Turbomachinery, ch. 1.

Question 2 of 8: Hydro Turbine Model

Question 2: Hydro Turbine Model (10 marks)