22-Mec-B6 Advanced Fluid Mechanics · December 2019
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
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 $15^{\circ}\text{C}$.
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 water power delivered by a pump is the product of three things, $P_{\text{hyd}} = \rho g Q H$, and on a constant-speed characteristic two of them move in opposite directions. At shut-off the valve is closed, the flow is zero, and the head is at its maximum; the product is therefore zero, because no water is being moved anywhere. At the other end of the curve — run-out, with the discharge fully open and the system offering no resistance — the flow is at its maximum but the head the pump can generate has collapsed to nearly nothing; the product is again zero, because although a great deal of water is moving it is being lifted no distance at all. Between these two zeros the product must rise to a maximum, and it does so somewhere near the middle of the flow range, a little to the right of the point of best efficiency.
The head–flow curve itself falls for a reason worth stating, because it is the reason the product has the shape it does. The Euler head an impeller with backward-curved vanes imparts is $H_{\text{Euler}} = \left(V_{B2}V_{2T} - V_{B1}V_{1T}\right)/g$, and since $V_{2T} = V_{B2} - V_{2R}/\tan\beta_2$ while $V_{2R}$ is proportional to flow, the ideal head falls linearly as the flow rises. Two loss families then bend that straight line into the curve actually measured. Friction losses in the impeller and volute grow roughly with the square of flow, pulling the curve down increasingly at high flow. Incidence (shock) losses are zero only at the one flow rate for which the blade inlet angle matches the relative flow direction — the shockless-entry condition examined quantitatively in Question 3 — and grow quadratically with the departure from it in either direction. The result is the familiar drooping head curve, and the water-power parabola that follows from multiplying it by $Q$.
The word "efficiency" in the question's part (a) is used loosely for the water horsepower curve; if the true efficiency $\eta = P_{\text{hyd}}/P_{\text{brake}}$ is meant, its shape has the same explanation with one extra ingredient. It is zero at shut-off, where the pump absorbs real power and delivers none; it is zero again at run-out, where it delivers no head; and in between it peaks at the best efficiency point, where the incidence loss vanishes and the friction loss has not yet become dominant. That peak is the duty the impeller was designed for, and it is the point at which the pump should be selected to run.
The difference between the brake horsepower absorbed at the coupling and the water horsepower delivered to the fluid is the total loss, and the shape of that gap follows from the fact that the two curves have completely different forms. Brake power does not start at zero. At shut-off the impeller is still spinning a body of water that is going nowhere: it churns it, recirculates it, heats it, and drags it round the volute, and it also has to overcome bearing friction, seal drag and disc friction on the faces of the impeller shrouds. That is the finite shut-off power at the left of the chart, and every watt of it is loss because no water is being delivered. The gap between the two curves is therefore at its widest, in relative terms, at zero flow.
As the valve opens, water power climbs steeply from zero while brake power rises only gently, so the gap closes. It reaches its minimum at, or very close to, the best efficiency point: the recirculation and churning of shut-off have disappeared because the flow now follows the blade passages properly, the incidence loss has fallen to zero because the relative flow enters along the blade, and friction losses are still modest because the velocities are moderate. This is simply the same statement as "efficiency is a maximum" viewed as a difference rather than a ratio.
Beyond the best efficiency point the gap opens again, and it opens faster than it closed. Three effects compound. Friction losses in the impeller and volute scale with the square of the velocities and therefore with the square of the flow. Incidence loss returns, now with the relative flow striking the pressure side of the blade instead of the suction side. And on a mixed-flow machine like the one charted, brake power keeps rising more or less linearly with flow while the head — and with it the water power — is collapsing towards zero. At run-out the pump is absorbing more shaft power than it did at shut-off and delivering nothing at all, so the gap exceeds its initial value. That is exactly the behaviour the question describes, and it is the reason a mixed-flow or axial pump must never be run far out on its curve: the motor can be overloaded and the machine will very likely cavitate as well, since the NPSH required rises steeply with flow.
The number of vanes controls how faithfully the water is forced to follow the blade shape, and both extremes are bad. The starting point is that the Euler equation assumes perfect guidance — that the water leaves the impeller exactly along the blade direction. A real impeller has a finite number of blades, so between two adjacent blades the water is only loosely constrained, and a relative eddy forms in the passage: because the passage rotates while the fluid within it tends to retain its orientation in the absolute frame, the flow circulates backwards relative to the blade. This reduces the exit whirl below the Euler value, and the ratio of actual to ideal whirl is called the slip factor.
Too few vanes means wide passages and weak guidance. The relative eddy is large, the slip factor is low — perhaps 0.7 or less — and the head developed falls well below what the blade geometry promises. The flow distribution across the passage is very non-uniform, with high velocity near the pressure side and low velocity, sometimes reversed, near the suction side, which increases mixing losses downstream at the volute. The pressure difference across each individual blade is large, which loads the blade heavily and makes the suction surface prone to local low pressure and hence to cavitation. The head–flow curve of such an impeller can also become unstable at low flow (a rising portion at small $Q$), which causes surging in a system with any storage capacity.
Too many vanes guides the flow beautifully — the slip factor approaches unity — but at a rising price. Each vane occupies part of the flow area with its thickness, so the effective throat area shrinks (the blockage factor falls) and the meridional velocity for a given flow rises. Each vane also adds wetted surface, so friction losses grow roughly in proportion to the number of blades. Crucially, the passages become long and narrow, with a large hydraulic-length-to-width ratio, and the friction loss in such a passage grows quickly. At the inlet eye the blockage is most damaging of all, because the increased velocity there depresses the local static pressure and worsens the NPSH required, so an over-bladed impeller cavitates earlier. The net effect is that the head gained from better guidance is more than cancelled by the head lost to friction and blockage, and efficiency falls.
The optimum therefore sits where the marginal gain in slip factor equals the marginal loss to friction and blockage, and for ordinary radial impellers this lands between five and eight vanes. Low-specific-speed radial impellers, whose passages are long and narrow and whose slip is worst, sit at the upper end; high-specific-speed mixed-flow impellers, whose short wide passages guide the flow well anyway and whose blockage penalty is severe, sit at the lower end with as few as three or four. A common empirical guide is $Z = 6.5\,\dfrac{D_2 + D_1}{D_2 - D_1}\sin\!\left(\dfrac{\beta_1 + \beta_2}{2}\right)$, which for the impeller of Question 3 — $D_1 = 130$ mm, $D_2 = 300$ mm, $\beta_1 = 20^{\circ}$, $\beta_2 = 25^{\circ}$ — returns $Z = 6.3$, so six vanes. A further practical refinement is the use of splitter vanes: short blades placed between the full-length ones and starting part way out from the eye, which restore guidance in the outer part of the passage where slip is generated without adding blockage at the inlet eye where cavitation is decided. Finally, the vane count is normally chosen so that it shares no common factor with the number of volute tongues or diffuser vanes, because coincident blade passing generates strong pressure pulsations and noise.
