22-Mec-B6 Advanced Fluid Mechanics · December 2018

Question 5 of 8: Curtis Type Impulse Turbine

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format: National Examinations, December 2018 — 16-Mec-B6 Fluid Machinery. Closed book, three hours, 22 pages. Section A is calculative (Questions 1–5) and Section B descriptive (Questions 6–8); candidates answer four questions from Section A and two from Section B — six questions of ten marks each, 60 marks in all. Reference data for individual questions are supplied on pages 11–17, and the nomenclature, general constants and reference equations on pages 18–22. All eight questions are solved below.

Check — angle conventions are taken from the paper's own attachments, and they are not the same across the paper. The compressor velocity diagram on page 11 strikes $\alpha_1$ and $\beta_1$ off the axial component $C_{X1}$, so every blade and vane angle in Questions 1 and 2 is measured from the axial direction. The steam-turbine diagram on page 16 strikes $\theta$, $\phi$, $\gamma$ and $\delta$ off the tangential (blade-motion) direction, and the Francis diagram on page 15 defines $\alpha_1$ between $V_1$ and $u_1$ — also tangential. Questions 4 and 5 therefore use the tangential reference. Every constant below is the paper's own page-19 value: $g = 9.81\ \text{m}\,\text{s}^{-2}$, $\rho_{\text{water}} = 1000\ \text{kg}\,\text{m}^{-3}$, $c_p = 1005\ \text{J}\,\text{kg}^{-1}\text{K}^{-1}$ and $c_v = 718\ \text{J}\,\text{kg}^{-1}\text{K}^{-1}$, from which $R = c_p - c_v = 287\ \text{J}\,\text{kg}^{-1}\text{K}^{-1}$ and $k = c_p/c_v = 1.3997$ — not a textbook 1.40.

Reference texts

Dixon, S. L. and Hall, C. A., Fluid Mechanics and Thermodynamics of Turbomachinery, 7th ed. — Ch. 5 (axial compressors), Ch. 4 (axial turbines), Ch. 9 (hydraulic turbines).
Cohen, H., Rogers, G. F. C. and Saravanamuttoo, H. I. H., Gas Turbine Theory, 7th ed. — Ch. 5 (axial flow compressors), Ch. 6 (compressor and turbine matching, surge and stall), Ch. 3 (jet propulsion cycles).
Turton, R. K., Principles of Turbomachinery, 2nd ed. — Ch. 2 (Euler equation), Ch. 8 (fans, blade form and control).
White, F. M., Fluid Mechanics, 8th ed. — Ch. 11 (turbomachinery: specific speed, Francis and Kaplan turbines, cavitation).
Fox, R. W. and McDonald, A. T., Introduction to Fluid Mechanics, 10th ed. — Ch. 10 (fluid machinery, similarity rules).
Çengel, Y. A. and Boles, M. A., Thermodynamics: An Engineering Approach, 9th ed. — Ch. 9 (gas power cycles, Brayton and turbojet).

Question 5 — Curtis Type 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.

Quantity	Symbol	Value
Absolute steam velocity leaving the nozzles	$V_{S1}$	$1411\ \text{m/s}$
Nozzle angle, from the plane of rotation	$\theta$	$20^{\circ}$
Number of moving rows (velocity compounding)	$n$	$2$
Fixed and moving blades	—	symmetrical in each row
Fluid friction in nozzles and blades	—	zero
Steam mass flow	$\dot M$	$100\ \text{kg/s}$

Find. The optimum blade speed, both combined velocity diagrams with every velocity and angle on them, the work of each row, the total power at 100 kg/s, and the blade (diagram) efficiency.

Approach. Fix the optimum blade speed for a two-row wheel, resolve the nozzle velocity into whirl and axial components, then walk through the four blade rows in turn — each symmetrical, frictionless row returns the relative velocity unchanged in magnitude with its whirl reversed — and take the work of each moving row from the change of whirl.

Optimum blade velocity (part a). For a velocity-compounded wheel with $n$ moving rows the work is maximised, and the exit kinetic energy minimised, when $$V_B = \frac{V_{S1}\cos\theta}{2n} = \frac{1411\cos 20^{\circ}}{2 \times 2} = \frac{1325.91}{4}$$ $$\boxed{V_B = 331.5\ \text{m/s}}$$ Using the single-row result $V_{S1}\cos\theta/2$ here is the single biggest trap in the question: it doubles the blade speed, leaves a large negative whirl at exit and destroys the axial discharge that defines the optimum.
Resolve the nozzle velocity. Because the page-16 attachment measures $\theta$ from the plane of rotation, the whirl is a cosine and the axial component a sine: $$V_{w1} = V_{S1}\cos 20^{\circ} = 1325.91\ \text{m/s},\qquad V_a = V_{S1}\sin 20^{\circ} = 482.59\ \text{m/s}$$ With no friction and symmetrical blades the axial component is the same everywhere in the machine, which is what allows every subsequent triangle to be closed without further data.
First moving row, inlet (part b and part c). Subtracting the blade speed from the whirl gives the relative velocity entering the first moving blades: $$\begin{aligned}V_{Rw1} &= V_{w1} - V_B = 1325.91 - 331.48 = 994.43\ \text{m/s} \\ V_{R1} &= \sqrt{994.43^2 + 482.59^2} = 1105.3\ \text{m/s}\end{aligned}$$ and the blade inlet angle, measured from the tangential direction, is $\phi = \arctan(482.59/994.43) = 25.9^{\circ}$.

Combined velocity diagram for the first moving row, drawn to the page-16 convention: nozzle velocity and relative inlet velocity above the blade-speed line, relative and absolute exit velocities below it. Reversing the relative whirl across a symmetrical frictionless blade is what turns 1411 m/s into 820 m/s of absolute exit velocity.

First moving row, outlet. The blades are symmetrical and frictionless, so the relative velocity leaves with the same magnitude and the same angle but with its whirl reversed. Adding the blade speed back gives the absolute exit condition: $$\begin{aligned}V_{w2} &= V_B - V_{Rw1} = 331.48 - 994.43 = -662.95\ \text{m/s} \\ V_{S2} &= \sqrt{662.95^2 + 482.59^2} = 820.0\ \text{m/s}\end{aligned}$$ The negative sign says the steam leaves the first row moving against the direction of blade motion, at $\delta = 143.9^{\circ}$ from the tangential, which is precisely the swirl the fixed row exists to turn round.
Fixed (guide) row. The fixed blades are symmetrical and frictionless, so they reverse the whirl without changing the magnitude. The steam therefore enters the second moving row at $V_{S3} = 820.0\ \text{m/s}$ with whirl $+662.95\ \text{m/s}$, at $33.6^{\circ}$ — more precisely $\arctan(482.59/662.95) = 36.1^{\circ}$ — from the tangential.
Second moving row (part b and part c concluded). Repeating the first-row construction with the reduced whirl: $$\begin{aligned}V_{Rw3} &= 662.95 - 331.48 = 331.48\ \text{m/s} \\ V_{R3} &= \sqrt{331.48^2 + 482.59^2} = 585.5\ \text{m/s}\end{aligned}$$ with blade inlet angle $\arctan(482.59/331.48) = 55.5^{\circ}$. Reversing the relative whirl once more and adding the blade speed gives $$V_{w4} = 331.48 - 331.48 = 0,\qquad V_{S4} = V_a = 482.6\ \text{m/s}$$ The steam leaves the wheel purely axially. That is the definition of the optimum blade speed, and its appearing here is the strongest possible confirmation that step 1 is right.

Combined diagram for the second moving row. The exit absolute velocity is vertical — purely axial, zero whirl — which is the geometric signature of the optimum blade speed and of the minimum leaving-loss the question asks for.

Work done by each row (part d). The page-21 relation makes the specific work the blade speed times the change of whirl across the row: $$w_1 = V_B\,(V_{w1}-V_{w2}) = 331.48\,[1325.91-(-662.95)] = 659.3\ \text{kJ/kg}$$ $$w_2 = V_B\,(V_{w3}-V_{w4}) = 331.48\,(662.95-0) = 219.8\ \text{kJ/kg}$$ $$\boxed{w_1 = 659.3\ \text{kJ/kg},\quad w_2 = 219.8\ \text{kJ/kg},\quad w_{\text{total}} = 879.0\ \text{kJ/kg}}$$ The rows split the work in the ratio 3 : 1, exactly the $(2n-1) : 1$ pattern a two-row Curtis wheel must show — a free audit that costs nothing to apply.
Total power (part e). At the stated steam flow, $$\boxed{P = \dot M\,w_{\text{total}} = 100 \times 879.0 = 87\,900\ \text{kW} = 87.9\ \text{MW}}$$
Blade efficiency (part f). Blade or diagram efficiency is the work done divided by the kinetic energy supplied by the nozzles: $$\eta_b = \frac{w_{\text{total}}}{V_{S1}^2/2} = \frac{879\,014}{1411^2/2} = \frac{879\,014}{995\,461}$$ $$\boxed{\eta_b = 0.883 = 88.3\ \text{per cent}}$$ For a frictionless wheel at the optimum speed this must equal $\cos^2\theta = \cos^2 20^{\circ} = 0.8830$, and it does to four figures. A second, independent check closes the energy books: the work plus the leaving kinetic energy, $879\,014 + 482.59^2/2 = 995\,461\ \text{J/kg}$, reproduces the nozzle kinetic energy exactly.

Quantity	Symbol	Result
Optimum blade velocity	$V_B$	$331.5\ \text{m/s}$
Axial velocity (constant throughout)	$V_a$	$482.6\ \text{m/s}$
Row 1: relative inlet velocity / angle	$V_{R1}$, $\phi_1$	$1105.3\ \text{m/s}$, $25.9^{\circ}$
Row 1: absolute exit velocity / angle	$V_{S2}$, $\delta$	$820.0\ \text{m/s}$, $143.9^{\circ}$
Fixed row exit (row 2 inlet)	$V_{S3}$	$820.0\ \text{m/s}$ at $36.1^{\circ}$
Row 2: relative inlet velocity / angle	$V_{R3}$, $\phi_2$	$585.5\ \text{m/s}$, $55.5^{\circ}$
Row 2: absolute exit velocity	$V_{S4}$	$482.6\ \text{m/s}$, purely axial
Work, first and second rows	$w_1$, $w_2$	$659.3$ / $219.8\ \text{kJ/kg}$ (3 : 1)
Total work	$w_{\text{total}}$	$879.0\ \text{kJ/kg}$
Total power at $100\ \text{kg/s}$	$P$	$87.9\ \text{MW}$
Blade efficiency	$\eta_b$	$88.3\ \text{per cent}$ ($=\cos^2\theta$)

Topic. Velocity compounding: the two-row Curtis impulse wheel, its optimum blade speed and its diagram efficiency.

Key relations. Optimum blade speed $V_B = V_{S1}\cos\theta/2n$; work per row $w = V_B\,\Delta V_w$; power $P = \dot M w$; blade efficiency $\eta_b = w/(V_{S1}^2/2)$, which equals $\cos^2\theta$ at the optimum for frictionless symmetrical blades; energy closure $w_{\text{total}} + V_{\text{exit}}^2/2 = V_{S1}^2/2$.

Why this works. An impulse wheel does work only by turning the steam, and the rate at which it does so is the blade speed times the change in the tangential momentum. A single row can absorb only a limited share of a very fast jet, because the best it can do is reverse the relative whirl, and the leaving kinetic energy remains large. Velocity compounding solves this by extracting the whirl in instalments: the first moving row takes a bite, a fixed row turns the residual swirl round without extracting any work, and a second moving row takes the rest. The optimum speed is the one that leaves the steam with no whirl at all after the last row, and dividing the single-row optimum by the number of rows is exactly the condition that achieves it. The work then falls in the ratio $3:1$ for two rows because each row sees a whirl change of $4V_B$ and $2V_B$ respectively. The whole construction depends on the axial component being conserved, which the frictionless-symmetrical-blade assumption guarantees.

Common pitfall. Taking $V_B = V_{S1}\cos\theta/2$ (the single-row optimum), which yields 663 m/s and a large negative exit swirl instead of an axial discharge. The second frequent error is measuring the nozzle angle from the axial direction, which swaps the sine and the cosine and destroys both the whirl and the work; the page-16 attachment shows $\theta$ struck off the blade-motion line, so the whirl is $V_{S1}\cos\theta$.

Source. Dixon & Hall, Fluid Mechanics and Thermodynamics of Turbomachinery, 7th ed., §4.2–4.4 (axial impulse stages, velocity compounding); Granet & Bluestein, Thermodynamics and Heat Power, 6th ed., Ch. 9 (steam turbine stages); White, Fluid Mechanics, 8th ed., §11.6.

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