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 distinction is about where the pressure falls. In a pure impulse stage the entire pressure drop of the stage occurs in the fixed (nozzle) row. The gas or steam leaves the nozzles as a fast jet and enters the moving blades at that speed; inside the moving passage the pressure is constant, so no further expansion occurs there. Because there is no expansion in the rotor, the passage between two moving blades has a constant flow area, and the velocity of the fluid relative to the blade leaves with the same magnitude it entered — only its direction has been changed. In a reaction stage, by contrast, the pressure drop is shared: part occurs in the fixed row and part in the moving row. The moving passage is convergent, the fluid expands as it passes through, and the relative velocity leaving the rotor is larger than the relative velocity entering it. The degree of reaction is defined as the fraction of the stage's static enthalpy drop that occurs in the rotor, so it is zero for a pure impulse stage and one half for the very common fifty per cent reaction design.
The forces are created in different ways, and this is what the question is really asking about. In the impulse rotor the only mechanism available is the change of momentum direction. The fluid enters the passage with a tangential momentum component and leaves with that component reversed; by Newton's second law applied to the control volume enclosing one passage, the blade must exert a force equal and opposite to that rate of change, and the reaction on the blade is what drives the wheel. The force is therefore purely a momentum effect, distributed as a pressure difference between the concave and convex sides of the blade but with equal static pressure at inlet and outlet. In the reaction rotor the same momentum-change force is present, but a second contribution is added: the static pressure at the passage inlet is higher than at its outlet, so there is a net pressure force on the blade in the direction of motion as well — the same effect that drives a rotating lawn sprinkler. The moving row is in effect a set of small nozzles carried on the wheel, and it is thrust backwards as the fluid is accelerated forwards out of them.
The consequences for the machine are systematic. An impulse stage can take a large pressure drop in one step, because the whole drop is handled in stationary nozzles where sealing is easy and no rotating seal has to hold a pressure difference; this makes impulse blading attractive for the high-pressure end of a steam turbine and for control (partial-admission) stages. But its blade efficiency is lower, because the high relative velocity through the rotor produces large friction losses, and the optimum blade speed ratio is only about one half of the nozzle-exit velocity. A reaction stage develops less work per stage for the same blade speed — roughly half as much, since the whirl change is smaller — and it needs a pressure-tight shroud or careful tip clearance control because the rotor now works against a pressure difference. In return it is aerodynamically cleaner: the accelerating flow keeps boundary layers thin in both rows, so stage efficiency is typically several points higher, and the optimum blade speed ratio rises to about unity, which suits the high blade speeds of a modern gas turbine. Most real machines are neither extreme: reaction is varied from hub to tip along each blade, so the root may be nearly impulse while the tip runs at forty or fifty per cent reaction.
Take a jet of constant velocity $V$ striking a bucket that recedes at velocity $U$. The water approaches the bucket at a relative velocity $V - U$; inside the bucket it is turned through nearly 180° and, in the ideal frictionless case, leaves with the same relative speed reversed. The whirl of the water therefore changes by $2(V - U)$ and the power delivered to the wheel is
$$P = \rho Q\,\Delta V_w\,U = 2\rho Q (V - U) U$$while the power available in the jet is $\tfrac{1}{2}\rho Q V^2$. The wheel efficiency is the ratio of these, which in terms of the speed ratio $\varphi = U/V$ is
$$\eta = \frac{2\rho Q (V-U)U}{\tfrac{1}{2}\rho Q V^{2}} = 4\varphi\left(1 - \varphi\right)$$a downward parabola through the origin and through $\varphi = 1$, with its maximum of unity at $\varphi = \tfrac{1}{2}$. Allowing for bucket friction (a relative-velocity retention factor $k$ of about 0.9) and for the fact that a real bucket turns the water through about 165° rather than 180°, so that the jets clear the following bucket, the expression generalises to $\eta = 2\varphi(1-\varphi)(1 + k\cos\beta)$, which is the same parabola scaled down to a peak of about 0.93.
The two ends of the curve are worth explaining separately, because that is what the question asks for. At zero blade velocity the bucket is stationary. The jet is turned through 180° and a large force is developed — in fact the maximum force the jet can exert, $2\rho Q V$ — but the point of application does not move, so no work is done. The water leaves with the same speed it arrived, carrying away all its kinetic energy, and the efficiency is zero. This is the stalled or locked-rotor condition, and it is why a Pelton wheel produces maximum torque at standstill and yet no power at all. At the other extreme, when the blade velocity equals the jet velocity, the bucket runs away from the water at exactly the speed the water is travelling. The relative velocity is zero, no water enters the bucket, no momentum is exchanged, no force is developed, and again no work is done — this time because the force has vanished rather than the displacement. This is the runaway condition of an unloaded wheel.
Between these two zeros the efficiency must rise and fall, and the parabola shows it peaking exactly halfway. The physical statement of the optimum is elegant: at $\varphi = \tfrac{1}{2}$ the water leaves the bucket with zero absolute velocity. Its relative speed on leaving is $V - U = \tfrac{1}{2}V$ directed backwards, and the bucket carries it forwards at $U = \tfrac{1}{2}V$; the two cancel exactly, so the water simply drops out of the bucket into the tailrace with no residual kinetic energy. Every joule the jet carried has been handed to the wheel. Above and below that ratio some kinetic energy always survives in the discharge — forwards if the wheel is running fast, backwards if it is running slow — and that residue is precisely the shortfall from unity.
Two practical remarks complete the picture. First, the curve is very flat near its peak: at $\varphi = 0.45$ or $0.55$ the ideal efficiency is still 0.99, which is why a Pelton unit can be locked to a synchronous speed that does not exactly suit the head and lose almost nothing, as the Bridge River figures in Question 5 demonstrate. Second, because the jet velocity is fixed by the head and the wheel speed is fixed by the grid, the designer's free variable is the pitch diameter, and choosing it to land near $\varphi = 0.46$ is one of the first decisions in laying out an impulse plant.
