Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format: National Examination 98-Mar-B5 Fluid Machinery, May 2017 — 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, cv,air = 0.718 kJ/kg·K.
Reference texts. S. L. Dixon & C. A. Hall, Fluid Mechanics and Thermodynamics of Turbomachinery (7th ed.); R. A. Sabersky, A. J. Acosta, E. G. Hauptmann & E. M. Gates, Fluid Flow: A First Course in Fluid Mechanics (4th ed.); H. Cohen, G. F. C. Rogers & H. I. H. Saravanamuttoo, Gas Turbine Theory; R. W. Fox, A. T. McDonald & P. J. Pritchard, Introduction to Fluid Mechanics.
Check: This paper, although listed under “Marine Control Systems”, is printed and headed “98-Mar-B5, Fluid Machinery” throughout, with zero marine-control-systems content (no PID loops, governors, or automation). It is solved here exactly as printed. Question 2's design method (annulus sizing from mass flow, 50% reaction symmetric velocity triangles) follows the standard preliminary axial-compressor design procedure of Dixon & Hall ch. 3–5; since no stage/compressor efficiency is supplied, the number of stages is obtained from the ideal (isentropic) overall temperature rise divided by the actual work done per stage — the standard simplification for this class of preliminary-design problem. Question 3 assumes zero exit whirl at the Francis runner outlet (design/best-efficiency condition), the standard assumption when no exit-blade data is given.
Approach. Invert the SI turbine specific speed $N_s=\omega\sqrt{P}/[\rho^{1/2}(gH)^{5/4}]$ (as given on the reference sheet) to get shaft power, use the hydraulic efficiency to get the flow rate, size the runner diameter from the tip-speed condition, then use Euler's equation (no exit whirl assumed) with the given inlet blade angle to get the radial velocity and, from continuity, the runner inlet height.
(a) Power output from specific speed. $\omega=2\pi N/60=2\pi(750)/60=78.54\ \text{rad/s}$; $gH=9.81(160)=1569.6\ \text{m}^2/\text{s}^2$. Solving $N_s=\omega\sqrt{P}/[\rho^{1/2}(gH)^{5/4}]$ for P,$$P=\left[\frac{N_s\,\rho^{1/2}(gH)^{5/4}}{\omega}\right]^2=\left[\frac{0.9\,(1000)^{1/2}(1569.6)^{1.25}}{78.54}\right]^2.$$$$\boxed{P=12.8\ \text{MW}}$$(Ns = 0.9 sits squarely in the Francis band of the attached efficiency-vs-specific-speed chart, 0.3–2.0, confirming the turbine type.)
(b) Water flow rate. The hydraulic efficiency relates shaft power to the available water power, $\eta_h=P/(\rho gQH)$, so$$Q=\frac{P}{\eta_h\,\rho g H}=\frac{12.82\times10^6}{0.94(1000)(9.81)(160)}.$$$$\boxed{Q=8.69\ \text{m}^3/\text{s}}$$
(c) Runner diameter. The free jet velocity for the head is $V_{jet}=\sqrt{2gH}=\sqrt{2(9.81)(160)}=56.03\ \text{m/s}$, so the runner tip speed is $U_1=0.7(56.03)=39.22\ \text{m/s}$. With $U_1=\pi D_1N/60$,$$D_1=\frac{60\,U_1}{\pi N}=\frac{60(39.22)}{\pi(750)}=0.999\ \text{m}.$$$$\boxed{D_1\approx1.00\ \text{m}}$$
(d) Inlet velocity diagram and radial velocity. Assuming no exit whirl (best-efficiency design condition, $V_{w2}=0$), Euler's equation gives the inlet whirl velocity directly from the hydraulic efficiency,$$V_{w1}=\frac{\eta_h\,gH}{U_1}=\frac{0.94(9.81)(160)}{39.22}=37.62\ \text{m/s}.$$Because $\alpha_1$ is measured from the tangent, the radial (meridional) component is $V_{f1}=V_{w1}\tan\alpha_1=37.62\tan18^{\circ}=12.22\ \text{m/s}$, and the absolute inlet velocity is $V_1=\sqrt{V_{w1}^2+V_{f1}^2}=39.56\ \text{m/s}$.$$\boxed{V_{f1}=12.2\ \text{m/s}}$$
Fig. Q3 - Francis runner inlet velocity triangle: blade speed U1, absolute velocity V1 (18° from tangent), and its radial component Vf1.
(e) Runner height at inlet. Continuity through the cylindrical inlet control surface (neglecting vane blockage) gives $Q=\pi D_1\,b_1\,V_{f1}$, so$$b_1=\frac{Q}{\pi D_1 V_{f1}}=\frac{8.69}{\pi(0.999)(12.22)}.$$$$\boxed{b_1=0.227\ \text{m}}$$
Check: The runner exit is assumed to carry no residual whirl (Vw2 = 0), the standard best-efficiency-point design target for a Francis runner discharging into a draft tube — the exit angle/geometry is not given, so this is the only tractable assumption. Vane thickness blockage is neglected in the inlet-height continuity calculation, consistent with a preliminary (first-pass) sizing exercise.
