22-Mec-A6 Fluid Machinery · December 2014
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Paper format: National Examination 07-Mec-A6-1 Fluid Machinery, December 2014 — 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.) — source of the pump-selection charts Figs 15.11/15.12; 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.
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.
In an impulse turbine the entire pressure (enthalpy) drop of the stage occurs in the fixed (nozzle) blades, which accelerate the fluid to a high absolute velocity. The moving blades then run at essentially constant pressure: they change only the direction of the relative velocity, whose magnitude is unchanged (apart from friction). The force on the blade is the pure momentum (impulse) reaction to this change of direction of the fast jet — hence the symmetric buckets of a Pelton wheel or the symmetric blades of a de Laval/Rateau stage.
In a reaction turbine the pressure drop is shared between the fixed and moving blades. The moving blades act as nozzles in their own right: the relative velocity increases across them as the pressure falls, and the blade force is produced partly by this acceleration (an aerodynamic lift/reaction, like the recoil of a nozzle) and partly by the change of direction. A 50% reaction stage splits the enthalpy drop equally and has mirror-image fixed and moving blade shapes. The two are contrasted in the velocity diagrams below: the impulse rotor keeps the relative speed constant, while the reaction rotor accelerates it.
Consequences: an impulse stage can take a large enthalpy drop per stage (fewer stages, but higher blade stresses and lower per-stage efficiency), and needs no pressure seal across the rotor. A reaction stage takes a smaller drop per stage (more stages, higher efficiency) but must handle the pressure difference across the rotor with tip sealing and axial-thrust balancing.
For a Pelton wheel with a constant jet velocity V, the work done per unit mass is w = U(V − U)(1 + k cosβ), where U is the bucket speed, β the bucket exit angle and k the friction factor. The diagram efficiency is this work divided by the jet kinetic energy ½V². Two limiting cases bound the behaviour: when U = 0 the buckets do not move, so although the force is maximum the distance moved (and hence the work) is zero; when U = V the buckets run away with the jet, the relative impact velocity is zero, no momentum is exchanged, and again no work is done. Between these, the product U(V − U) is a maximum at U = V/2, so the efficiency rises from zero, peaks near a blade-speed ratio of about 0.5, and falls back to zero — the parabolic curve below. In practice the peak sits slightly below 0.5 (≈ 0.46–0.47, as used in Question 1) because of bucket friction and windage.
