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 2015 — 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).
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. Design head, flow and geometric ratios for an impulse (Pelton) wheel.
Given data
Head, H
200 m
Flow, Q
4 m³/s
Nozzle velocity coefficient, K
0.99
Wheel diameter, D
1.47 m
Mechanical efficiency
88 %
Speed ratio U/V ; jet/wheel d/D
0.47 ; 0.113
Find. Rotational speed, shaft power, nozzle count and specific speed.
Figure 2. Multi-jet Pelton wheel: jets strike the buckets at the pitch circle; blade speed U = 0.47 V.
Approach. Get the jet velocity from the head and nozzle coefficient, set the blade speed from the ratio to fix rotational speed, size a single jet to count the nozzles, and evaluate the shaft power and specific speed.
Wheel speed. $U=0.47V=29.1\ \text{m/s}$, and $U=\dfrac{\pi D N}{60}$ gives $N=\dfrac{60U}{\pi D}=\dfrac{60\cdot29.1}{\pi\cdot1.47}=\boxed{379\ \text{rev/min}}$.
Power output. With nozzle and bucket losses taken as ideal, the mechanical efficiency represents the water-to-shaft conversion: $P=\eta_m\,\rho g Q H=0.88\cdot1000\cdot9.81\cdot4\cdot200=6.91\ \text{MW}$.
Number of nozzles. Jet diameter $d=0.113\cdot1.47=0.166\ \text{m}$, so one jet passes $q=\tfrac{\pi}{4}d^{2}V=1.34\ \text{m}^3/\text{s}$; the number of jets is $n=Q/q=4/1.34=2.98\Rightarrow3$ nozzles.
Specific speed. $\Omega_{sp}=\dfrac{\omega P^{1/2}}{\rho^{1/2}(gH)^{5/4}}=0.25$ (equivalently a dimensional $N_s\approx42$ in kW-m units), squarely in the Pelton range.
Check: no bucket deflection angle or blade-friction factor is supplied, so nozzle and bucket (hydraulic) losses are taken as ideal and the stated 88 % mechanical efficiency is applied as the overall water-to-shaft efficiency. If a bucket angle of about 165° were assumed instead, the wheel efficiency would be near unity and the power would change by only a few percent.
The number of nozzles falling almost exactly on 3 is the internal check that the jet velocity and jet diameter are consistent: three jets of 0.166 m each carry the 4 m³/s design flow. Multi-jetting is precisely how a Pelton wheel raises its specific speed — each added jet multiplies the flow at the same head and speed, moving the machine from the low-specific-speed single-jet regime toward the Francis range.
