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25-Nav-B5 Marine Control Systems · May 2015

Question 4 of 8: Curtis (Velocity-Compounded) Impulse Turbine

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.

It is solved here exactly as printed.

Question 4: Curtis (Velocity-Compounded) Impulse Turbine (10 marks)

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.

Given. A two-row Curtis stage with a single set of nozzles feeding both moving rows.

Given data
Nozzle exit (steam) velocity, Vs11411 m/s
Nozzle angle20°
Bladessymmetrical, frictionless
Rows (velocity-compounded)2
Steam flow100 kg/s

Find. Optimum blade speed, all velocity-triangle quantities, work per stage, total power and blade efficiency.

U = 331.5 m/sVr1=1105Vs1=1411 (20°)Stage-1 inlet triangle (scale 1 m/s = 0.30 mm shown)
Figure 3. First-stage inlet velocity triangle (blade speed U along the wheel, absolute jet at 20°).

Approach. For a velocity-compounded wheel with $n$ rows the optimum blade speed is $U=\tfrac{V_{s1}\cos\alpha}{2n}$; build symmetric, frictionless triangles row by row, then sum the Euler work.

  1. Optimum blade speed. With $n=2$, $U=\dfrac{V_{s1}\cos20^{\circ}}{2\cdot2}=\dfrac{1411\cdot0.9397}{4}=\boxed{331.5\ \text{m/s}}$.
  2. Stage-1 relative velocity and angle. Whirl $V_{w1}=1411\cos20^{\circ}=1325.9$ and flow $V_{f}=1411\sin20^{\circ}=482.6\ \text{m/s}$ give $V_{r1}=\sqrt{(V_{w1}-U)^{2}+V_f^{2}}=1105\ \text{m/s}$ at $\beta_1=25.9^{\circ}$.
  3. Stage-1 work. Symmetric frictionless blades reverse the relative whirl, so the change in absolute whirl is large: $w_1=U\,\Delta V_{w}=659.3\ \text{kJ/kg}$.
  4. Stage-2 work. The fixed row redirects the flow into the second moving row; carrying the same construction gives $w_2=219.8\ \text{kJ/kg}$ — the classic 3 : 1 split of a two-row Curtis stage.
  5. Total work and power. $w=w_1+w_2=879\ \text{kJ/kg}$; for 100 kg/s, $P=879\cdot100=87.9\ \text{MW}$.
  6. Blade efficiency. $\eta_b=\dfrac{w}{\tfrac12 V_{s1}^{2}}=0.883=\cos^{2}20^{\circ}$ — the exact ideal result.

The blade efficiency collapsing to $\cos^{2}\alpha$ is the elegant self-check of the ideal velocity-compounded stage: for symmetric frictionless blading at the optimum speed the whole two-row wheel converts exactly $\cos^{2}20^{\circ}=88.3\%$ of the jet kinetic energy to work, independent of the individual row angles. The 3 : 1 work split shows why Curtis staging is used only where a large single pressure drop must be absorbed at low rotational speed: the second row does far less work than the first.

Question 4 results
QuantityValue
(a) Optimum blade speed331.5 m/s
(b) Velocity diagramFigure 3 (stage-1 inlet triangle, to scale)
(c) Vr1 / β11105 m/s / 25.9°
(d) Work stage 1 / stage 2659.3 / 219.8 kJ/kg
(e) Total power (100 kg/s)87.9 MW
(f) Blade efficiency88.3 %