22-Mec-B6 Advanced Fluid Mechanics · December 2019

Question 3 of 8: Pump Performance

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

Notes on this paper

Paper format: National Examinations, December 2019 — 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); a candidate answers 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 in the Attachments on pages 12–17, and the nomenclature, general constants and reference equations on pages 18–22. All eight questions are solved below, because this set is a study resource rather than a three-hour sitting.

Check — angle conventions are taken from the paper's own attachments, and on this sitting they are uniform. The gas-turbine velocity diagram on page 13 strikes $\theta$, $\phi$, $\gamma$ and $\delta$ off the tangential direction (the plane of rotation); the pump diagram on page 14 strikes $\alpha_1$, $\alpha_2$, $\beta_1$ and $\beta_2$ off the tangential direction as well; and the Francis diagram on page 15 defines $\alpha_1$ between $V_1$ and $u_1$, again tangential. Every angle below is therefore measured from the direction of blade motion. Every constant used is the paper's own page-19 value: $g = 9.81\ \text{m}\,\text{s}^{-2}$, $\rho_{\text{water}} = 1000\ \text{kg}\,\text{m}^{-3}$, $p_{\text{atm}} = 100\ \text{kPa}$ and $p_{\text{vapour}} = 1.71\ \text{kPa}$ at $15^{\circ}\text{C}$.

Reference texts

Dixon, S. L. and Hall, C. A., Fluid Mechanics and Thermodynamics of Turbomachinery, 7th ed. — Ch. 1 (specific speed), Ch. 4 (axial turbines), Ch. 7 (centrifugal pumps), Ch. 9 (hydraulic turbines, Thoma cavitation parameter). The efficiency chart on page 15 and the cavitation chart on page 17 of the examination paper are reproduced from this text.
Cohen, H., Rogers, G. F. C. and Saravanamuttoo, H. I. H., Gas Turbine Theory, 7th ed. — Ch. 5 (axial compressors, stage loading and stall), Ch. 7 (axial turbines, stage work and blade angles).
Turton, R. K., Principles of Turbomachinery, 2nd ed. — Ch. 2 (the Euler equation), Ch. 5 (radial machines, vane number and slip), Ch. 6 (hydraulic turbines).
White, F. M., Fluid Mechanics, 8th ed. — Ch. 11 (turbomachinery: pump characteristics, specific speed, Pelton, Francis and Kaplan turbines, cavitation). The pump characteristic curves reproduced on page 11 of the examination paper come from this text.
Fox, R. W., McDonald, A. T. and Pritchard, P. J., Introduction to Fluid Mechanics, 10th ed. — Ch. 10 (fluid machinery, Euler turbomachine equation, cavitation and NPSH).
Douglas, J. F., Gasiorek, J. M. and Swaffield, J. A., Fluid Mechanics, 6th ed. — Ch. 23 (rotodynamic machines, impeller velocity triangles).

Question 3 — Pump Performance (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 centrifugal water pump whose impeller geometry, speed and duty are completely specified, with both blade angles struck from the tangential direction as the page-14 attachment defines them.

Quantity	Symbol	Value
Blade inner / outer diameter	$D_1$, $D_2$	$0.130$ / $0.300\ \text{m}$
Blade inner / outer height (axial)	$h_1$, $h_2$	$0.020$ / $0.010\ \text{m}$
Blade inlet / outlet angle (from the tangent)	$\beta_1$, $\beta_2$	$20^{\circ}$ / $25^{\circ}$
Pump speed	$N$	$1750\ \text{rev}\,\text{min}^{-1}$
Water flow rate	$Q$	$0.030\ \text{m}^3\,\text{s}^{-1}$
Hydraulic head delivered	$H$	$35\ \text{m}$
Density of water (page 19)	$\rho$	$1000\ \text{kg}\,\text{m}^{-3}$

Find. The blade speeds, the radial and tangential water velocities at inlet and outlet, the torque and shaft power the impeller absorbs, and the hydraulic power delivered together with the resulting efficiency.

Approach. Take the two velocity triangles in turn. The blade speed comes from the rotational speed and the radius; the radial velocity comes from continuity through the cylindrical periphery, not through a disc; the tangential water velocity then follows by closing the triangle against the known blade angle. Torque comes from the moment-of-momentum equation, and efficiency from comparing the head actually delivered with the head the impeller works for.

Tangential blade velocities (part a). The impeller is a rigid body, so the blade speed at any radius is simply $\pi D N/60$: $$V_{B1} = \frac{\pi \times 0.130 \times 1750}{60} = 11.91\ \text{m}\,\text{s}^{-1},\qquad V_{B2} = \frac{\pi \times 0.300 \times 1750}{60} = 27.49\ \text{m}\,\text{s}^{-1}$$ $$\boxed{V_{B1} = 11.91\ \text{m}\,\text{s}^{-1},\qquad V_{B2} = 27.49\ \text{m}\,\text{s}^{-1}}$$
Radial water velocities (part b). The water crosses a cylindrical surface of circumference $\pi D$ and height $h$, so the flow area is $\pi D h$ — never $\tfrac{\pi}{4}D^2$, which is the area of a disc and has nothing to do with a radial machine: $$V_{1R} = \frac{Q}{\pi D_1 h_1} = \frac{0.030}{\pi \times 0.130 \times 0.020} = 3.67\ \text{m}\,\text{s}^{-1}$$ $$V_{2R} = \frac{Q}{\pi D_2 h_2} = \frac{0.030}{\pi \times 0.300 \times 0.010} = 3.18\ \text{m}\,\text{s}^{-1}$$ The two are deliberately similar: the designer has narrowed the passage from 20 mm to 10 mm precisely to hold the meridional velocity roughly constant as the radius more than doubles.
Tangential water velocities (part c). With $\beta$ measured from the tangent, the relative velocity has tangential component $V_B - V_T$ and radial component $V_R$, so $\tan\beta = V_R/(V_B - V_T)$ and hence $$V_T = V_B - \frac{V_R}{\tan\beta}$$ At inlet and outlet respectively, $$V_{1T} = 11.91 - \frac{3.67}{\tan 20^{\circ}} = 11.91 - 10.09 = 1.82\ \text{m}\,\text{s}^{-1}$$ $$V_{2T} = 27.49 - \frac{3.18}{\tan 25^{\circ}} = 27.49 - 6.83 = 20.66\ \text{m}\,\text{s}^{-1}$$ $$\boxed{V_{1T} = 1.82\ \text{m}\,\text{s}^{-1},\qquad V_{2T} = 20.66\ \text{m}\,\text{s}^{-1}}$$ The absolute velocities that go with them are $V_1 = 4.10\ \text{m}\,\text{s}^{-1}$ at $\alpha_1 = 63.6^{\circ}$ and $V_2 = 20.91\ \text{m}\,\text{s}^{-1}$ at $\alpha_2 = 8.8^{\circ}$ from the tangent.

Impeller velocity triangles drawn to a common scale. The inlet triangle is short and steep because the blade speed is small and the water enters almost radially; the outlet triangle is long and shallow because the blade speed has more than doubled while the meridional velocity has barely changed. The blue vectors are the absolute water velocities, the green vectors the velocities relative to the blade.

Check — this impeller is not running at its shockless-entry duty, and the small inlet pre-whirl is real, not an arithmetic slip. A blade set for zero inlet whirl at this flow would need $\beta_1 = \arctan(V_{1R}/V_{B1}) = \arctan(3.67/11.91) = 17.1^{\circ}$. The blade is actually set at $20^{\circ}$, so at 0.030 m³/s the water must arrive with a small positive whirl of 1.82 m/s. That is normal — a pump has one shockless flow rate and is run over a range around it — but it means the standard textbook shortcut $V_{1T} = 0$ must not be applied here. Using it would raise the computed torque by about 1.3 per cent and, more importantly, would discard the information the question is testing in part (c).

Torque on the impeller (part d). The page-22 hydraulic torque is the rate of change of moment of momentum between the two radii: $$\tau = \rho Q\left(r_2 V_{2T} - r_1 V_{1T}\right) = 1000 \times 0.030 \times \left(0.150 \times 20.66 - 0.065 \times 1.82\right)$$ $$\tau = 30 \times \left(3.099 - 0.118\right) = 89.4\ \text{N}\,\text{m}$$
Power required to drive the impeller (part d, continued). With $\omega = 2\pi N/60 = 183.3\ \text{rad}\,\text{s}^{-1}$, $$\boxed{\tau = 89.4\ \text{N}\,\text{m},\qquad P_{\text{shaft}} = \tau\omega = 89.4 \times 183.3 = 16.39\ \text{kW}}$$ This is the power the water absorbs from the blades; mechanical losses in bearings and seals would add a little more at the coupling, but the question asks only for the impeller.
Cross-check through the Euler head. The same power must follow from the Euler pump equation, which is a useful independent route: $$H_{\text{Euler}} = \frac{V_{B2}V_{2T} - V_{B1}V_{1T}}{g} = \frac{27.49 \times 20.66 - 11.91 \times 1.82}{9.81} = 55.70\ \text{m}$$ and $\rho g Q H_{\text{Euler}} = 1000 \times 9.81 \times 0.030 \times 55.70 = 16.39\ \text{kW}$, which reproduces the torque route exactly. The Euler head is the head the blades impart; the 35 m the pump actually delivers is what survives after hydraulic losses.
Hydraulic power and efficiency (part e). The useful output is the water power at the stated head: $$P_{\text{hyd}} = \rho g Q H = 1000 \times 9.81 \times 0.030 \times 35 = 10.30\ \text{kW}$$ so the efficiency of the pump is $$\boxed{P_{\text{hyd}} = 10.30\ \text{kW},\qquad \eta = \frac{10.30}{16.39} = 0.629 = 62.9\ \text{per cent}}$$ The 20.7 m gap between the Euler head and the delivered head — friction in the passages, incidence loss at the 20° inlet blade, slip at the impeller exit, and mixing in the volute — is the whole of the loss. Sixty-three per cent is a believable, if unremarkable, figure for a small single-stage pump running off its best-efficiency point, which is exactly what the inlet-incidence check above showed it to be doing.

Question 3 — final results
Quantity	Symbol	Result
(a) Blade velocity at inlet / outlet	$V_{B1}$, $V_{B2}$	$11.91$ / $27.49\ \text{m}\,\text{s}^{-1}$
(b) Radial water velocity at inlet / outlet	$V_{1R}$, $V_{2R}$	$3.67$ / $3.18\ \text{m}\,\text{s}^{-1}$
(c) Tangential water velocity at inlet / outlet	$V_{1T}$, $V_{2T}$	$1.82$ / $20.66\ \text{m}\,\text{s}^{-1}$
(c) Absolute water velocity at inlet / outlet	$V_1$, $V_2$	$4.10$ / $20.91\ \text{m}\,\text{s}^{-1}$
(d) Torque required to drive the impeller	$\tau$	$89.4\ \text{N}\,\text{m}$
(d) Power required to drive the impeller	$P_{\text{shaft}}$	$16.39\ \text{kW}$
(d) Euler head imparted by the blades	$H_{\text{Euler}}$	$55.70\ \text{m}$
(e) Hydraulic (water) power	$P_{\text{hyd}}$	$10.30\ \text{kW}$
(e) Pump efficiency	$\eta$	$62.9\ \text{per cent}$

Topic. Centrifugal pump performance from impeller velocity triangles and the Euler turbomachine equation.

Key relations. Blade speed, $V_B = \pi D N/60$; continuity through a cylindrical periphery, $V_R = Q/(\pi D h)$; triangle closure with $\beta$ from the tangent, $V_T = V_B - V_R/\tan\beta$; hydraulic torque, $\tau = \rho Q (r_2 V_{2T} - r_1 V_{1T})$; power, $P = \tau\omega$; Euler head, $gH_{\text{Euler}} = V_{B2}V_{2T} - V_{B1}V_{1T}$; water power, $P_{\text{hyd}} = \rho g Q H$.

Why this works. Everything a rotodynamic machine does to a fluid is contained in one statement: the torque on the shaft equals the rate at which the fluid's moment of momentum changes between inlet and outlet. Dividing that torque by the mass flow and multiplying by the angular velocity turns it into the Euler head, which is the ideal head the blading imparts. The triangles are the bookkeeping that lets you get at the two whirl velocities, and the only geometric input they need is the flow area — which for a radial machine is the curved surface $\pi D h$, because the water leaves radially through the rim rather than axially through a face. Comparing the Euler head with the head actually measured then separates the machine's two quite different loss families: everything hydraulic shows up as the difference between those two heads, while bearing and seal losses show up only between the coupling and the impeller.

Common pitfall. Using $\tfrac{\pi}{4}D^2$ for the flow area is the classic error and inflates $V_{2R}$ by a factor of about seven here. The second trap is assuming $V_{1T} = 0$ because "pumps are designed for shockless entry": that is a design condition, not a given, and this impeller is demonstrably not at it. A third is measuring $\beta_2$ from $V_{2T}$ rather than from $V_{B2} - V_{2T}$, which silently returns the absolute flow angle $\alpha_2$ instead of the blade angle.

Source. Dixon & Hall, Fluid Mechanics and Thermodynamics of Turbomachinery, 7th ed., §7.2–7.5 (centrifugal pump, inlet and outlet velocity triangles, slip and efficiency); Douglas, Gasiorek & Swaffield, Fluid Mechanics, 6th ed., §23.3–23.5; White, Fluid Mechanics, 8th ed., §11.2 (Euler turbomachine equations).

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