25-Nav-B5 Marine Control Systems · May 2018
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Paper format: National Examinations, May 2018 — 98-Mar-B5 Fluid Machinery. Closed book, three hours, 21 pages. Section A is calculative (Questions 1–5) and Section B descriptive (Questions 6–8); candidates answer four from Section A and two from Section B, six questions of ten marks each for a total of 60. Reference data for individual questions are supplied on pages 11–16 and the nomenclature, constants and reference equations on pages 17–21. All eight questions are solved below, because the set is a study resource rather than a three-hour sitting.
Check — angle conventions taken from the paper's own attachments. The compressor attachment on page 11 strikes $\alpha_1$ and $\beta_1$ off the axial component $C_{X1}$, so in Questions 1 and 2 all blade and vane angles are measured from the axial direction. The steam-turbine attachment on page 13 and the Francis attachment on page 14 strike $\theta$, $\phi$, $\gamma$, $\delta$, $\alpha$ and $\beta$ off the tangential (blade-motion) direction, so Questions 3 and 5 use the tangential reference. Mixing the two is the single most common way to lose all the marks on a velocity-diagram question. Every constant used below ($g = 9.81\ \text{m}\,\text{s}^{-2}$, $\rho_{\text{water}} = 1000\ \text{kg}\,\text{m}^{-3}$, $\rho_{\text{air}} = 1.21\ \text{kg}\,\text{m}^{-3}$ at 15 °C, $c_p = 1.005$ and $c_v = 0.718\ \text{kJ}\,\text{kg}^{-1}\text{K}^{-1}$) is taken from the paper's page 18 rather than from a textbook.
Each five-mark part below is answered at the length the paper asks for: a full page of explanation with supporting sketches.
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 blade speed of any rotating blade row is not a single number: it is $U = \omega r$, so it grows in direct proportion to radius from the hub to the tip. The meridional (axial) velocity, by contrast, is very nearly uniform across the span, because it is set by continuity through the annulus — the mass flow divided by the annulus area and the density. A blade therefore sees a relative flow whose tangential component varies enormously along its length while its axial component barely changes at all, and the relative flow angle $\beta = \arctan\left[(U - V_\theta)/V_x\right]$ swings correspondingly.
The first stage of the Question 1 compressor makes the point numerically. The root runs at 160 m/s and the tip at 320 m/s, while the axial velocity is 204.5 m/s everywhere and the guide vanes deliver 36.06 m/s of whirl at every radius. The relative inlet angle is therefore 31.2° at the root and 54.2° at the tip — a swing of 23° across a blade only 92 mm tall. On a large low-pressure steam-turbine blade, where the tip radius can be three or four times the root radius, the swing routinely exceeds 40°.
An untwisted blade can only be set at one angle, so it can only be aligned with the incoming relative flow at one radius. Everywhere else the flow arrives at a large positive or negative incidence. Positive incidence at the hub drives the suction surface towards separation and, in a compressor, towards stall; negative incidence at the tip loads the pressure surface, thickens the wake and can choke the passage. Both extremes generate large profile and secondary losses, they lower the stage efficiency, and in a compressor the hub separation is the seed of rotating stall and surge. Twisting the blade — setting a progressively different metal angle at each radius, so that the leading edge is aligned with the local relative flow all the way up the span — restores shock-free (or near-shock-free) entry everywhere.
There is a second, equally important reason. In a machine with long blades the flow is not in simple axial equilibrium: the swirl imposed on the fluid produces a centrifugal force $\rho V_\theta^2/r$ that must be balanced by a radial pressure gradient. The resulting radial-equilibrium condition ties the spanwise distribution of whirl to the spanwise distribution of axial velocity. Only certain whirl distributions give a well-behaved, radially stable flow; the classical choice is the free vortex, $V_\theta r = \text{constant}$, which delivers constant axial velocity and constant work at every radius. But a free-vortex whirl distribution demands a different inlet and outlet angle at every radius, which is precisely a twisted blade. Blades on small machines (short blades, hub-to-tip ratio near unity) can be untwisted because neither effect is significant over their span; the ratio of tip to root radius, not the absolute size, is what forces the twist.
The two triangles above are drawn with a common axial velocity and a common absolute inlet whirl, differing only in blade speed. At the base the blade speed is small, so the relative velocity vector lies close to the axial direction and its angle to the axis is modest; the blade section there is steeply staggered and heavily cambered, because it must turn a nearly axial relative flow through a large deflection. At the tip the blade speed is large, the relative velocity is dominated by its tangential component and lies close to the plane of rotation; the section is nearly flat to the tangential direction and needs much less camber, because the deflection required of it is small.
Three consequences follow directly from the shape of the two triangles, and they are what a sketch is being asked to show. First, the relative velocity magnitude grows strongly towards the tip — here from 205 m/s at the root to 350 m/s at the tip — which is why tip Mach number, not root Mach number, limits the design. Second, the absolute exit velocity from a turbine blade is smallest where the whirl is best removed, and on a long blade the whirl distribution cannot be optimised at every radius with an untwisted section. Third, the work done per unit mass, $U\Delta V_\theta$, would rise steeply with radius if $\Delta V_\theta$ were held constant, which is exactly what the free-vortex design avoids by making $\Delta V_\theta \propto 1/r$; that distribution is achieved by the twist.
Definition. The degree of reaction $R$ of a turbine (or compressor) stage is the fraction of the stage's total static enthalpy change that occurs in the moving blade row:
$$R = \frac{\text{static enthalpy drop in the rotor}}{\text{total static enthalpy drop in the stage}} = \frac{h_1 - h_2}{h_0 - h_2}$$Because static enthalpy and static pressure move together in a nozzle or a blade passage, $R$ is equivalently the fraction of the pressure drop taken in the rotor. $R = 0$ is a pure impulse stage: all the expansion happens in the fixed nozzles, the steam crosses the moving blades at constant pressure, and the blades merely deflect it — exactly the arrangement analysed in Question 3, where $V_{R2} = V_{R1}$. $R = 0.5$ is the classical fifty-percent reaction stage, in which fixed and moving rows share the drop equally and the two blade rows have identical (mirror-image) profiles. Values above 0.5 put most of the expansion in the rotor. In terms of the velocity diagram, for constant axial velocity the reaction can be written as
$$R = 1 - \frac{V_{\theta 1} + V_{\theta 2}}{2U}$$which shows immediately why $R$ must vary along a long blade: the numerator is a property of the swirl the fixed row imposes, while the denominator is the blade speed, which rises linearly with radius.
How $R$ changes from base to tip. Under free-vortex design the whirl varies as $V_\theta \propto 1/r$, so the reaction takes the form $R = 1 - K/r^2$: it is smallest at the hub and largest at the tip, and it rises steeply because the radius appears squared. Taking the Question 1 first stage as an illustration and imposing $R = 0.5$ at mid-height, the same free-vortex distribution gives $R = -0.13$ at the root and $R = 0.72$ at the tip. On a large low-pressure steam turbine, where the tip-to-root radius ratio is far greater, designers typically arrange the blade so that the root is at or near zero reaction (impulse) and the tip reaches 0.5 or above.
Three practical consequences follow. At the root, low or zero reaction means little pressure drop across the moving row, which is desirable because it minimises the leakage flow through the diaphragm gland and avoids a large axial thrust; but it also means the relative velocity is high and the passage is heavily loaded, which is why root sections are thick and strongly cambered. At the tip, high reaction means a substantial pressure drop across the rotating row, so tip leakage over the shroud becomes a real loss and shrouding or tip seals are needed. If the reaction at the root is allowed to go appreciably negative, the static pressure would rise through the rotor at that radius — a diffusing relative flow in a turbine, which invites separation and reverse flow at the hub. Avoiding that condition is one of the practical limits on how long a blade can be for a given hub radius, and it is a common reason for departing from a strict free vortex in favour of a constant-reaction or controlled-vortex design.
