Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format: National Examination 98-Mar-B5 Fluid Machinery, December 2016 — closed book, three hours, 60 marks. Section A is calculative (Q1–Q5) and Section B is descriptive (Q6–Q8); the rubric asks for four questions of Section A plus two of Section B (six questions, each of equal value, 10 marks). All eight questions are solved in full as a study resource. General constants supplied with the paper: g = 9.81 m/s², patm = 100 kPa, pvapour = 2.34 kPa (20 °C), ρwater = 1000 kg/m³, ρair = 1.21 kg/m³ (15 °C), cp,air = 1.005 kJ/kg°C, k = 1.4, R = 0.287 kJ/kg K.
Reference texts. S. L. Dixon & C. A. Hall, Fluid Mechanics and Thermodynamics of Turbomachinery, 7th ed.; R. K. Turton, Principles of Turbomachinery, 2nd ed.; H. Cohen, G. F. C. Rogers & H. I. H. Saravanamuttoo, Gas Turbine Theory, 6th ed.; R. W. Fox, A. T. McDonald & P. J. Pritchard, Introduction to Fluid Mechanics, 8th ed.
Given. Francis turbine, electrical output $P_e=200\ \text{MW}$, net head $H=110\ \text{m}$, dimensionless power specific speed $\Omega_{sp}=0.9$, $\alpha_1=22^\circ$, $V_{abs}/V_{jet}=0.77$, $V_{blade}/V_{jet}=0.6583$, with $V_{jet}=\sqrt{2gH}$.
Find. Turbine efficiency and hydraulic input power, rotational speed and runner diameter, the runner-inlet velocity triangle and radial velocity, and the flow rate and runner-inlet height.
Fig. 3.1 — Runner-inlet velocity triangle: absolute velocity V1 at 22° to the tangent (blade speed u1), resolved into whirl Vw1 and radial Vr1.
Approach. Read $\eta$ from the Francis efficiency chart to get hydraulic power; invert the dimensionless specific speed for $\omega$; get the blade speed from the velocity ratio to size the runner; resolve the inlet triangle for the radial velocity; and close with continuity for the flow and inlet height.
Efficiency and hydraulic power (a). From the Page-13 chart, the Francis line at $\Omega_{sp}=0.9$ reads $\eta\approx0.94$. Hence $$P_{hyd}=\frac{P_e}{\eta}=\frac{200}{0.94}=\boxed{212.8\ \text{MW}}$$
Rotational speed (b). From $\Omega_{sp}=\dfrac{\omega\sqrt{P}}{\rho^{1/2}(gH)^{5/4}}$ with $P=P_e$: $$\omega=\frac{\Omega_{sp}\,\rho^{1/2}(gH)^{5/4}}{\sqrt{P}}=\frac{0.9\times31.62\times(1079.1)^{1.25}}{\sqrt{200\times10^6}}=12.45\ \text{rad/s}$$ $$N=\frac{60\omega}{2\pi}=\boxed{118.8\ \text{rev/min}}$$
Inlet velocity diagram & radial velocity (c). Absolute inlet velocity $V_1=0.77V_{jet}=35.77\ \text{m/s}$ at $\alpha_1=22^\circ$ to the tangent: $$V_{r1}=V_1\sin\alpha_1=35.77\sin22^\circ=\boxed{13.4\ \text{m/s}},\qquad V_{w1}=V_1\cos22^\circ=33.2\ \text{m/s}$$
Flow rate and inlet height (d). From $P_{hyd}=\rho g Q H$: $$Q=\frac{P_{hyd}}{\rho g H}=\frac{212.8\times10^6}{1000(9.81)(110)}=\boxed{197\ \text{m}^3/\text{s}}$$ and with $Q=\pi D\,b\,V_{r1}$ over the runner periphery, $$b=\frac{Q}{\pi D V_{r1}}=\frac{197}{\pi(4.91)(13.4)}=\boxed{0.95\ \text{m}}$$
