Question 1 of 8: Hydro Turbine Model (Vanderkloof)
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Examinations — 16-Mec-A6 Fluid Machinery, May 2019. Closed-book, three hours. Section A is calculative (5 questions) and Section B is descriptive (3 questions); candidates answer four questions from Section A and two from Section B (six questions, 60 marks, 10 marks each). All eight questions are solved as a study resource.
Reference texts. Dixon & Hall, Fluid Mechanics and Thermodynamics of Turbomachinery (7th ed.); Cohen, Rogers & Saravanamuttoo, Gas Turbine Theory (6th ed.); Fox & McDonald, Introduction to Fluid Mechanics (10th ed.); F. M. White, Fluid Mechanics (8th ed.); Çengel & Boles, Thermodynamics: An Engineering Approach (9th ed.).
Paper. This is the 16-Mec-A6, May 2019 Fluid Machinery examination.
Question 1: Hydro Turbine Model (Vanderkloof) (10 marks)
Given. Prototype Vanderkloof Francis unit: electrical output $P=120\;\text{MW}$, $N=125\;\text{rev/min}$, net head $H=65\;\text{m}$, design flow $Q=200\;\text{m}^3/\text{s}$, runner diameter $D_p=5.462\;\text{m}$. Homologous model: $D_m=0.200\;\text{m}$, test head $H_m=10\;\text{m}$.
Given data
Quantity
Prototype
Model
Runner diameter
5.462 m
0.200 m
Net head
65 m
10 m
Design flow
200 m³/s
— (find)
Speed
125 rev/min
— (find)
Electrical output
120 MW
— (find, ideal)
Find. Dimensionless specific speed (machine type), overall efficiency, and the model speed, flow, ideal power and Moody-scaled efficiency.
Approach. Classify the runner from the dimensionless power specific speed; get overall efficiency from the ratio of electrical output to hydraulic power $\rho g Q H$; scale the model with the head and flow coefficients $\dfrac{gH}{N^2D^2}$ and $\dfrac{Q}{ND^3}$; and correct efficiency for scale with the Moody equation.
Check: the “0.90 power factor” describes the generator's electrical loading and does not enter the hydraulic calculation — the 120 MW is real (active) power.
Dimensionless power specific speed. With $\omega=\dfrac{2\pi N}{60}=\dfrac{2\pi(125)}{60}=13.09\;\text{rad/s}$,
$$\Omega_{sp}=\frac{\omega\,\sqrt{P/\rho}}{(gH)^{5/4}}=\frac{13.09\sqrt{120\times10^{6}/1000}}{\left(9.81\times65\right)^{1.25}}=\boxed{1.42}$$
A value near 1.4 places the machine firmly in the Francis (radial/mixed-flow) range, consistent with a 65 m head.
Overall efficiency. The available hydraulic power is $\rho g Q H$, so
$$\eta_o=\frac{P}{\rho g Q H}=\frac{120\times10^{6}}{1000\times9.81\times200\times65}=\boxed{0.941\;(94.1\%)}$$
Model speed — head coefficient. Dynamic similarity keeps $\dfrac{gH}{N^2D^2}$ constant, hence
$$N_m=N_p\,\frac{D_p}{D_m}\sqrt{\frac{H_m}{H_p}}=125\times\frac{5.462}{0.200}\sqrt{\frac{10}{65}}=\boxed{1339\;\text{rev/min}}$$
Model flow — flow coefficient. Keeping $\dfrac{Q}{ND^3}$ constant,
$$Q_m=Q_p\,\frac{N_m}{N_p}\left(\frac{D_m}{D_p}\right)^{3}=200\times\frac{1339}{125}\left(\frac{0.200}{5.462}\right)^{3}=\boxed{0.105\;\text{m}^3/\text{s}\;(105\;\text{L/s})}$$
Ideal model power. With no losses the power equals the hydraulic power at the model head,
$$P_m=\rho g Q_m H_m=1000\times9.81\times0.105\times10=\boxed{10.3\;\text{kW}}$$
Moody efficiency scaling. First strip the generator to get the prototype hydraulic efficiency, $\eta_{h,p}=\dfrac{\eta_o}{\eta_{gen}}=\dfrac{0.941}{0.98}=0.960$. The Moody equation (head-inclusive form) gives the efficiency to demand of the smaller, higher-loss model:
$$1-\eta_m=\frac{1-\eta_{h,p}}{\left(\dfrac{D_m}{D_p}\right)^{0.25}\left(\dfrac{H_m}{H_p}\right)^{0.10}}\;\Rightarrow\;\eta_m=\boxed{0.890\;(89.0\%)}$$
The diameter-only form $1-\eta_m=(1-\eta_{h,p})\left(\tfrac{D_m}{D_p}\right)^{0.2}$ gives 92.3%; the head-inclusive value is the conservative target because the model runs at a much lower head.
