Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format: National Examinations, May 2018 — 98-Mar-B5 Fluid Machinery. Closed book, three hours, 21 pages. Section A is calculative (Questions 1–5) and Section B descriptive (Questions 6–8); candidates answer four from Section A and two from Section B, six questions of ten marks each for a total of 60. Reference data for individual questions are supplied on pages 11–16 and the nomenclature, constants and reference equations on pages 17–21. All eight questions are solved below, because the set is a study resource rather than a three-hour sitting.
Check — angle conventions taken from the paper's own attachments. The compressor attachment on page 11 strikes $\alpha_1$ and $\beta_1$ off the axial component $C_{X1}$, so in Questions 1 and 2 all blade and vane angles are measured from the axial direction. The steam-turbine attachment on page 13 and the Francis attachment on page 14 strike $\theta$, $\phi$, $\gamma$, $\delta$, $\alpha$ and $\beta$ off the tangential (blade-motion) direction, so Questions 3 and 5 use the tangential reference. Mixing the two is the single most common way to lose all the marks on a velocity-diagram question. Every constant used below ($g = 9.81\ \text{m}\,\text{s}^{-2}$, $\rho_{\text{water}} = 1000\ \text{kg}\,\text{m}^{-3}$, $\rho_{\text{air}} = 1.21\ \text{kg}\,\text{m}^{-3}$ at 15 °C, $c_p = 1.005$ and $c_v = 0.718\ \text{kJ}\,\text{kg}^{-1}\text{K}^{-1}$) is taken from the paper's page 18 rather than from a textbook.
Reference texts
Dixon, S. L. and Hall, C. A., Fluid Mechanics and Thermodynamics of Turbomachinery, 7th ed. — Ch. 3 (two-dimensional cascades), Ch. 5 (axial compressors), Ch. 4 (axial turbines), Ch. 9 (hydraulic turbines).
Cohen, H., Rogers, G. F. C. and Saravanamuttoo, H. I. H., Gas Turbine Theory, 7th ed. — Ch. 3 (jet propulsion cycles), Ch. 5 (axial flow compressors), Ch. 6 (compressor and turbine matching, surge and stall).
Turton, R. K., Principles of Turbomachinery, 2nd ed. — Ch. 2 (Euler equation), Ch. 4 (pump characteristics), Ch. 8 (fans and their control).
White, F. M., Fluid Mechanics, 8th ed. — Ch. 11 (turbomachinery: specific speed, pump curves, Pelton and Francis turbines).
Fox, R. W. and McDonald, A. T., Introduction to Fluid Mechanics, 10th ed. — Ch. 10 (fluid machinery, similarity rules).
Çengel, Y. A. and Boles, M. A., Thermodynamics: An Engineering Approach, 9th ed. — Ch. 9 (gas power cycles, Brayton and turbojet).
Subject note. Although this paper is listed under “Marine Control Systems”, page 1 of the examination itself reads 98-MAR-B5, FLUID MACHINERY, and every question is a turbomachine question with zero marine-control-systems content. The solutions below answer the paper as printed, citing turbomachinery texts rather than any control-systems reference.
Given. A single-jet Pelton turbine of dimensionless specific speed 0.20 running at 880 rev/min on an effective head of 120 m, with a nozzle velocity coefficient of 0.985, a bucket-to-jet speed ratio of 0.47 and an overall (shaft) efficiency of 0.88.
Given data (Question 4)
Turbine specific speed (dimensionless)
$N_s = 0.20$
Effective head at nozzle inlet
$H = 120\ \text{m}$
Nozzle velocity coefficient
$C_v = 0.985$
Runner rotational speed
$N = 880\ \text{rev}\,\text{min}^{-1}$
Blade-speed to jet-speed ratio
$\varphi = 0.47$
Overall efficiency (shaft output)
$\eta_o = 0.88$
Find. The shaft power, the volume flow rate, the jet flow area (and hence jet diameter), and the wheel-to-jet diameter ratio.
Figure 4.1 — Single-jet Pelton arrangement. The jet strikes the buckets tangentially on the pitch circle of diameter $D$; the diameter ratio $D/d$ measures how many jet diameters fit round the wheel and so governs bucket spacing and size.
Approach. Invert the specific-speed definition on the aid sheet for the shaft power, convert power to flow through the overall efficiency, get jet velocity from the nozzle coefficient and hence the jet area, then find the wheel diameter from the speed ratio.
Convert the rotational speed to angular velocity. The specific-speed definition on page 20 uses radians per second:
$$\omega = \frac{2\pi N}{60} = \frac{2\pi \times 880}{60} = 92.15\ \text{rad}\,\text{s}^{-1}$$
Part (a) — invert the specific-speed definition for shaft power. The aid sheet gives the turbine specific speed as $N_s = \omega P^{1/2}/\left[\rho^{1/2}(gH)^{5/4}\right]$, so
$$P = \left[\frac{N_s\,\rho^{1/2}\,(gH)^{5/4}}{\omega}\right]^{2} = \left[\frac{0.20 \times 31.62 \times (9.81 \times 120)^{1.25}}{92.15}\right]^{2}$$
$$\boxed{P_{\text{shaft}} = 224.0\ \text{kW}}$$
Substituting back, $\omega\sqrt{P}/[\rho^{1/2}(gH)^{5/4}] = 0.200$, recovering the given specific speed. A value of 0.20 sits squarely in the Pelton band (single-jet impulse machines run below about 0.3), which is a useful confirmation that the machine type and the data are consistent.
Part (b) — volume flow rate from the overall efficiency. The overall efficiency relates shaft output to the water power supplied, $P_{\text{shaft}} = \eta_o\rho gQH$, hence
$$Q = \frac{P_{\text{shaft}}}{\eta_o\rho gH} = \frac{223\,956}{0.88 \times 1000 \times 9.81 \times 120}$$
$$\boxed{Q = 0.2162\ \text{m}^{3}\,\text{s}^{-1}}$$
Checking, $0.88 \times 1000 \times 9.81 \times 0.2162 \times 120 = 224.0\ \text{kW}$. The water power delivered to the nozzle is $\rho gQH = 254.5\ \text{kW}$, of which 30.5 kW is lost in the nozzle, buckets, windage and bearings.
Part (c) — jet velocity and flow area. The nozzle converts the head into velocity with the stated coefficient:
$$V_{\text{jet}} = C_v\sqrt{2gH} = 0.985\sqrt{2 \times 9.81 \times 120} = 0.985 \times 48.52 = 47.79\ \text{m}\,\text{s}^{-1}$$
Continuity at the nozzle exit then gives the area, and hence the jet diameter,
$$A_j = \frac{Q}{V_{\text{jet}}} = \frac{0.2162}{47.79}$$
$$\boxed{A_j = 4.523 \times 10^{-3}\ \text{m}^{2} = 45.23\ \text{cm}^{2}, \qquad d_j = 75.89\ \text{mm}}$$
Part (d) — wheel diameter and the diameter ratio. The bucket speed follows from the given speed ratio, and the wheel diameter from the rotational speed:
$$U = \varphi V_{\text{jet}} = 0.47 \times 47.79 = 22.46\ \text{m}\,\text{s}^{-1}, \qquad D = \frac{60U}{\pi N} = \frac{60 \times 22.46}{\pi \times 880} = 0.4875\ \text{m}$$
Dividing by the jet diameter,
$$\boxed{\frac{D}{d_j} = \frac{0.4875}{0.07589} = 6.42}$$
A ratio of about six is at the low end of practical Pelton design (typical machines run from roughly 6 to 20). It is the direct consequence of the low specific speed being combined with a comparatively high rotational speed for direct generator drive: the wheel is small while the jet is relatively fat.
Check — assumed interpretation. The specific speed is taken as the dimensionless turbine form printed on the paper's page 20, $N_s = \omega P^{1/2}/[\rho^{1/2}(gH)^{5/4}]$, with $\omega$ in rad/s and SI units throughout. Had a dimensional definition been intended the numerical value 0.20 would be meaningless, so this reading is forced by the data. The "effective head at the nozzle inlet" is treated as the net head available for conversion in the nozzle, so it appears both in $V_{\text{jet}} = C_v\sqrt{2gH}$ and in the water power $\rho gQH$.
