Question 4 of 8: Hydro Turbine Model (Vanderkloof)
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format: National Examination 07-Mec-A6-1 Fluid Machinery, December 2016 — closed book, 3 hours. Section A (Calculative, Q1–Q5) and Section B (Descriptive, Q6–Q8); candidates do four of A and two of B for 60 marks. All eight questions are solved as a study resource. Values marked “from the figure” are read from the examination attachment drawings.
Reference texts: Dixon & Hall, Fluid Mechanics and Thermodynamics of Turbomachinery (7th ed.); Cohen, Rogers & Saravanamuttoo, Gas Turbine Theory (6th ed.); Turton, Principles of Turbomachinery; Çengel & Boles, Thermodynamics (9th ed.); Fox & McDonald, Introduction to Fluid Mechanics (9th ed.). Constants from the paper: $g=9.81\ \text{m/s}^2$, $c_p=1.005\ \text{kJ/kg\,K}$, $k=1.4$, $R=0.287\ \text{kJ/kg\,K}$, $\rho_{water}=1000\ \text{kg/m}^3$, $p_{atm}=100\ \text{kPa}$, $p_{vap}=2.34\ \text{kPa}$.
Question 4: Hydro Turbine Model (Vanderkloof) (10 marks)
Find. Prototype specific speed and overall efficiency; model speed, flow and ideal power; and the model efficiency required by the Moody scaling law.
Approach. Use the dimensionless specific speed and $\eta=P_e/\rho gQH$; scale the model with the head and flow affinity coefficients; and apply the Moody relation after stripping the generator (electrical) efficiency to isolate the hydraulic efficiency that scales.
Specific speed (a). $\omega=2\pi(125)/60=13.09\ \text{rad/s}$: $$\Omega_{sp}=\frac{\omega\sqrt{P_e}}{\rho^{1/2}(gH)^{5/4}}=\frac{13.09\sqrt{120\times10^6}}{31.62\,(637.65)^{1.25}}=\boxed{1.42}$$ (dimensionless — a mid-range Francis machine).
Overall efficiency (b). $$\eta=\frac{P_e}{\rho g Q H}=\frac{120\times10^6}{1000(9.81)(200)(65)}=\boxed{0.941}$$
Model speed (c). Equal head coefficient $gH/(N^2D^2)$: $$N_m=N_p\frac{D_p}{D_m}\sqrt{\frac{H_m}{H_p}}=125\times\frac{5.462}{0.20}\sqrt{\frac{10}{65}}=\boxed{1339\ \text{rev/min}}$$
Model flow (d). Equal flow coefficient $Q/(ND^3)$: $$Q_m=Q_p\frac{N_m}{N_p}\left(\frac{D_m}{D_p}\right)^3=200\times\frac{1339}{125}\left(\frac{0.20}{5.462}\right)^3=\boxed{0.105\ \text{m}^3/\text{s}}$$
Ideal model power (e). $$P_m=\rho g Q_m H_m=1000(9.81)(0.105)(10)=\boxed{10.3\ \text{kW}}$$
Model efficiency by Moody (f). Strip the 98% generator efficiency to get the prototype hydraulic efficiency $\eta_{h,p}=0.941/0.98=0.960$. The head-inclusive Moody relation $1-\eta_m=(1-\eta_p)\big[(D_m/D_p)^{1/4}(H_m/H_p)^{1/10}\big]^{-1}$ gives $$\eta_m=1-(1-0.960)\big[(0.0366)^{0.25}(0.1538)^{0.10}\big]^{-1}=\boxed{0.890}$$ The diameter-only approximation $(1-\eta_p)/(1-\eta_m)=(D_m/D_p)^{1/5}$ gives $\eta_m=0.923$; the model, being far smaller, is measurably less efficient than the prototype.
Check (which efficiency scales): The Moody law scales the hydraulic efficiency (size / Reynolds-dependent losses), not the generator. The 98% electrical efficiency is given precisely so it can be divided out before scaling; the model test measures hydraulic efficiency, so 0.890 (head-inclusive form) is the target reading. Because the model and prototype heads differ (10 m vs 65 m), the head-inclusive form is preferred over the diameter-only approximation.
