22-Mec-A6 Fluid Machinery · May 2016
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Paper format. National Examination 07-Mec-A6-1 Fluid Machinery (May 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 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. K. Turton, Principles of Turbomachinery; R. H. Sabersky, A. J. Acosta, E. G. Hauptmann & E. M. Gates, Fluid Flow: A First Course in Fluid Mechanics (4th ed.) — pump-characteristic and cavitation-parameter charts; R. W. Fox, A. T. McDonald & P. J. Pritchard, Introduction to Fluid Mechanics (pump energy equation and affinity laws); Y. A. Çengel & J. M. Cimbala, Fluid Mechanics: Fundamentals and Applications.
Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.
The distinction lies in where the pressure drop, and therefore the acceleration of the fluid, takes place. In an impulse turbine the entire stage pressure drop occurs in the fixed nozzles: the fluid emerges as a high-velocity jet, and the moving blades merely turn that jet, changing its direction while the pressure — and the magnitude of the relative velocity — stays essentially constant across the rotor. The force on the blades is then purely the reaction to the change of fluid momentum (the change of whirl), $F=\dot m\,\Delta V_w$, exactly Newton's second law applied to the deflected jet.
In a reaction turbine the pressure drops in both the fixed and the moving rows. The moving passages act as nozzles in their own right, so the relative velocity increases across the rotor. The blade force now has two contributions: an impulse part from turning the flow, and a reaction part from the fluid accelerating as it leaves the blade — the same principle that drives a rotating lawn sprinkler. The degree of reaction R is the fraction of the stage enthalpy drop that occurs in the moving row (R = 0 for pure impulse, R = 0.5 for the common 50%-reaction stage).
In velocity-diagram terms, the impulse stage has $|W_1|=|W_2|$ (relative speed unchanged, the triangle merely reflected), whereas the 50%-reaction stage has mirror-image fixed and moving rows, with $C_1=W_2$ and $W_1=C_2$: the absolute velocity rises in the fixed row and falls in the rotor, and the relative velocity does the opposite. Impulse staging tolerates a large drop per stage (fewer, robust stages); reaction staging gives higher efficiency but needs more stages and tight tip sealing because of the pressure drop across the rotor.
For a Pelton wheel with constant jet velocity V and blade velocity U, the diagram efficiency follows $\eta=2\dfrac{U}{V}\left(1-\dfrac{U}{V}\right)(1+k\cos\beta)$, a downward parabola in U/V that is zero at both ends and peaks near U/V = 0.5.
At U/V = 0 the wheel is stationary: the jet exerts a force on the buckets but, with no displacement, no work is done and η = 0. At U/V = 1 the buckets run away with the jet: the relative velocity approaching the bucket falls to zero, so there is no change of momentum, no force, and again no work — η = 0. Between these limits the product of force (which falls as U rises) and blade speed (which rises) is maximised when the bucket moves at half the jet speed, so that the water is turned through nearly 180° and leaves the bucket almost at rest in the fixed frame, surrendering nearly all of its kinetic energy. Friction and the finite bucket exit angle move the practical optimum a little below 0.5 (typically 0.46–0.47, as found in Question 3).
