25-Nav-B5 Marine Control Systems · May 2016
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Paper format: National Examination 98-Mar-B5 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 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. K. Turton, Principles of Turbomachinery, 2nd ed.; H. Cohen, G. F. C. Rogers & H. I. H. Saravanamuttoo, Gas Turbine Theory, 6th ed.; 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, 8th ed.; 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.
In a pressure-compounded impulse turbine the total pressure drop is divided among several stages, each stage consisting of a fixed nozzle row followed by a moving row. The pressure therefore falls as a descending staircase: it drops in each fixed (nozzle) row, where the steam is accelerated, and stays constant across each moving row, where the impulse blades only turn the flow. The absolute velocity does the opposite — it rises sharply in each nozzle row (pressure energy converted to kinetic energy) and falls in each moving row as the steam gives up its kinetic energy to the blades. Over four stages this produces a repeated saw-tooth in velocity riding on a stepwise-descending pressure line.
(a) Why the blade lengthens and twists. As the steam expands through the turbine its pressure falls and its specific volume increases by orders of magnitude. To pass the same mass flow at a roughly constant axial velocity, the flow annulus area — and therefore the blade height — must grow steadily toward the low-pressure exhaust end. Once a blade is long, the blade speed $U=\omega r$ varies substantially from root to tip, because r changes appreciably over the blade height while ω is fixed. If the blade angle were constant, the incidence would be wrong everywhere except one radius, causing shock losses and uneven work. The blade is therefore twisted (a free-vortex design), its inlet and outlet angles changing continuously from root to tip so the relative flow enters each section smoothly and the work per unit mass stays reasonably uniform along the span.
(b) Root, mid, tip velocity diagrams. Keeping the axial component and the swirl schedule of the free-vortex design, the blade speed grows from root to tip: (i) at the root U is smallest, so the relative velocity W carries the most whirl, lying most inclined to the axial direction (smallest blade angle measured from the plane of rotation); (ii) at mid-length U is larger, W swings closer to axial and the blade angle opens out; (iii) at the tip U is largest, W is nearly axial and the blade angle approaches 90° from the plane of rotation (the blade section is nearly straight and set almost axially) — the three diagrams below show this progression.
