25-Nav-B5 Marine Control Systems · December 2017
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Paper format. 98-Mar-B5 Fluid Machinery, National Examinations December 2017 — three hours, closed book. Section A is calculative (Questions 1–5) and Section B descriptive (Questions 6–8); the rubric asks for four of Section A and two of Section B, six questions of ten marks each for a sixty-mark paper. Reference data for individual questions are supplied as Attachments (pages 10–17) and a general nomenclature/constants/equations sheet occupies pages 18–22. All eight questions are solved here, because the set is a study resource rather than a timed attempt.
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.; F. M. White, Fluid Mechanics, 8th ed.; R. W. Fox, A. T. McDonald & P. J. Pritchard, Introduction to Fluid Mechanics, 9th ed.; Y. A. Çengel & M. A. Boles, Thermodynamics: An Engineering Approach, 9th ed. Constants are those printed on page 19 of the paper (g = 9.81 m/s², ρwater = 1000 kg/m³, ρair = 1.21 kg/m³ at 15 °C and 1.19 kg/m³ at 20 °C, cp = 1.005 kJ/kg°C, cv = 0.718 kJ/kg°C, patm = 100 kPa, pvapour = 1.71 kPa at 15 °C). Every reference equation quoted below is one of those printed on pages 20–22, and is identified as such where it is first used.
The paper notes that “each five mark part of each question requires a full page answer with complete explanations with sketches, if appropriate, to support the explanation.” The three answers below are written to that instruction: argued prose, with a sketch wherever the physics is easier to show than to say.
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 between an impulse and a reaction turbine is a question of where the pressure drop is taken, and everything else follows from that single choice.
In an impulse stage the whole of the available pressure drop occurs in the fixed nozzles. The working fluid enters the nozzle at high pressure and low velocity and leaves it at the pressure that will exist throughout the moving row, converted almost entirely into kinetic energy. The moving blades then form a passage of constant flow area, symmetrical about its own mid-plane. Across that passage the static pressure does not change, so in the frame of the moving blade the relative velocity is unchanged in magnitude — it is only turned. The blade force is therefore purely the rate of change of momentum of the jet as it is deflected, which is what “impulse” means. Because the pressure is the same on both sides of the moving row, there is no axial pressure thrust on the rotor, and no incentive for the fluid to leak around the blade tips; that is why an impulse row can be built with generous tip clearance and, at the extreme, why a Pelton wheel can run in open air.
In a reaction stage the pressure drop is shared between the fixed and the moving rows — equally, in the fifty-per-cent design that is almost universal in steam and gas turbines. The fixed row now behaves as a partial nozzle, accelerating the fluid and turning it, and the moving row is itself a convergent passage. In the rotating frame the relative velocity therefore accelerates through the moving blades as well as being turned. The blade force now has two parts: the impulse force from the change of direction, and an additional reaction force that arises because the fluid is expanding and accelerating within the blade passage, exactly as a rocket nozzle pushes on the vehicle that carries it. The consequence is a net static pressure difference across the moving row, which produces a substantial axial thrust that the bearings (or a balance piston, or opposed flow paths) must absorb, and which drives leakage over the blade tips — so reaction blading needs shrouding and tight clearances.
The velocity diagrams make the difference immediate. For a symmetrical impulse row the inlet and outlet relative velocity vectors have the same length, so the diagram is a mirror pair about the axial direction, and the whole of the whirl change — and so the whole of the work — comes from deflection alone. For a fifty-per-cent reaction row the two triangles are congruent but interchanged: the relative outlet triangle is the mirror image of the absolute inlet triangle, so that W2 = C1 and C2 = W1, and the fixed and moving blades can be made to the same profile. Half the enthalpy drop is delivered in each row.
The engineering consequence is the choice of blade speed. Because an impulse row must absorb the whole of the stage's kinetic energy in one deflection, its optimum blade-speed ratio is U/C1 = cosα/2 — a fast wheel for a given enthalpy drop, or equivalently a large enthalpy drop for a given wheel speed. A reaction row only has to absorb half, and its optimum is U/C1 = cosα, twice as high, meaning half the enthalpy drop per stage at the same rim speed. Reaction machines therefore need roughly twice as many stages as impulse machines, but they are more efficient stage for stage because the accelerating passage keeps the boundary layer attached and the mean velocity level lower. The classic compromise — an impulse (often Curtis) first stage followed by reaction stages — takes the large first pressure drop where a robust, thrust-free, partial-admission wheel does the job well, then hands over to reaction blading for the bulk of the expansion, where efficiency matters most.
A Pelton wheel is the purest impulse machine: a free jet of speed V strikes a bucket moving at speed U in the same direction, is turned through nearly 180°, and leaves. The jet speed is fixed by the head and does not care what the wheel is doing, so the only variable is U.
Work out the two factors separately. In the frame of the bucket the water arrives at the relative speed V − U, and (neglecting friction) leaves at the same relative speed, reversed. The change of tangential momentum per unit mass is therefore (1 + kcosβ)(V − U), where β is the small deflection short of a full reversal and k a bucket friction factor. The force on the bucket is proportional to that, and the power is the force times the bucket speed:
$$P=\dot m\,(1+k\cos\beta)(V-U)\,U$$The two factors move in opposite directions as U is raised, and that opposition is the entire explanation the question asks for.
At U = 0 the wheel is stalled against the brake. The jet strikes the stationary bucket with the greatest possible relative speed, so the force is at its maximum — but the bucket does not move, no distance is covered, and no work whatever is done. The efficiency is zero and all the jet energy is dissipated in the splash. At U = V the opposite happens: the buckets run away from the jet at exactly the jet's own speed, the water never catches up, the relative velocity and hence the force are zero, and again no work is done. The efficiency is zero for a second and quite different reason.
Between those two zeros the product (V − U)U is a downward parabola with its maximum where the derivative vanishes, at U = V/2. Physically that is the condition at which the water leaves the bucket with almost no absolute velocity at all: it arrives at V, is reversed to a relative −(V − U) = −V/2, and adding the bucket speed V/2 back gives an absolute exit velocity of zero. All of the jet's kinetic energy has been surrendered to the wheel, which is why the ideal efficiency at that point reaches (1 + kcosβ)/2 — close to unity for a well-formed bucket. In practice the peak is measured at about U/V = 0.46 rather than 0.50, because windage, bearing friction and the need to clear the outgoing water from the path of the following bucket all grow with wheel speed, and because a real deflection of about 165° leaves a small residual velocity. The curve is flat near its crest, so a Pelton wheel tolerates a wide speed range with little efficiency penalty — a useful property for a machine that must hold synchronous speed while the head varies.
