Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
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.
Reference texts
Dixon, S. L. and Hall, C. A., Fluid Mechanics and Thermodynamics of Turbomachinery, 7th ed. — Ch. 3 (two-dimensional cascades), 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. 3 (jet propulsion cycles), Ch. 5 (axial flow compressors), Ch. 6 (compressor and turbine matching, surge and stall).
Turton, R. K., Principles of Turbomachinery, 2nd ed. — Ch. 2 (Euler equation), Ch. 4 (pump characteristics), Ch. 8 (fans and their control).
White, F. M., Fluid Mechanics, 8th ed. — Ch. 11 (turbomachinery: specific speed, pump curves, Pelton and Francis turbines).
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).
Subject note. Although this paper is listed under “Marine Control Systems”, page 1 of the examination itself reads 98-MAR-B5, FLUID MACHINERY, and every question is a turbomachine question with zero marine-control-systems content. The solutions below answer the paper as printed, citing turbomachinery texts rather than any control-systems reference.
Given. The first stage of an eight-stage axial compressor running at 16 500 rpm passes 20 kg/s of air drawn in at 15 °C, with a tip speed of 320 m/s and a hub-to-tip diameter ratio of one half.
Given data (Question 1)
Rotational speed
$N = 16\,500\ \text{rev}\,\text{min}^{-1}$
Blade tip velocity
$U_{\text{tip}} = 320\ \text{m}\,\text{s}^{-1}$
Root/tip diameter ratio (stage 1)
$D_{\text{root}} = 0.5\,D_{\text{tip}}$
Air mass flow rate
$\dot{M} = 20\ \text{kg}\,\text{s}^{-1}$
Air inlet temperature
$T_1 = 15\ ^\circ\text{C} = 288.15\ \text{K}$
Air density at 15 °C (page 18)
$\rho = 1.21\ \text{kg}\,\text{m}^{-3}$
Guide-vane outlet angle (from axial)
$\alpha_1 = 10^\circ$
Rotor blade outlet angle (from axial)
$\beta_2 = 30^\circ$
Find. The first-stage root and tip diameters, the axial inlet velocity, the mean blade speed, the complete inlet and outlet velocity triangles at mid-height, and the power absorbed by the first stage.
Figure 1.1 — First-stage velocity triangles at mid-height. The axial velocity $C_X$ is common to both triangles, so both share the same vertical height, and the two tangential components on each baseline must add to the blade speed $U$.
Approach. Fix the annulus geometry from the tip speed, get the axial velocity from continuity through that annulus, evaluate the blade speed at mid-height, then close the two velocity triangles with a constant axial velocity and apply the Euler work equation $w = U(C_{Y2} - C_{Y1})$.
Part (a)(i) — size the first-stage annulus from the tip speed. The tip velocity fixes the tip diameter directly:
$$D_{\text{tip}} = \frac{60\,U_{\text{tip}}}{\pi N} = \frac{60 \times 320}{\pi \times 16\,500}$$
which gives $D_{\text{tip}} = 0.3704\ \text{m}$, and with the stated hub-to-tip ratio $D_{\text{root}} = 0.5 \times 0.3704 = 0.1852\ \text{m}$. Checking backwards, $\pi \times 0.3704 \times 16\,500/60 = 320\ \text{m}\,\text{s}^{-1}$ as required, so
$$\boxed{D_{\text{tip}} = 0.370\ \text{m} = 370.4\ \text{mm}, \qquad D_{\text{root}} = 0.185\ \text{m} = 185.2\ \text{mm}}$$
Part (a)(ii) — obtain the annulus flow area. Neglecting blade thickness, the whole annulus is open to the flow:
$$A_1 = \frac{\pi}{4}\left(D_{\text{tip}}^{2} - D_{\text{root}}^{2}\right) = \frac{\pi}{4}\left(0.3704^{2} - 0.1852^{2}\right) = 0.08081\ \text{m}^{2}$$
Because the root is exactly half the tip, three-quarters of the swept tip circle is open area — a useful arithmetic check.
Part (a)(ii) concluded — apply continuity for the axial inlet velocity. At inlet the flow has no significant compressibility correction, so with the paper's own inlet density $\rho = 1.21\ \text{kg}\,\text{m}^{-3}$,
$$C_X = \frac{\dot{M}}{\rho A_1} = \frac{20}{1.21 \times 0.08081}$$
$$\boxed{C_X = 204.5\ \text{m}\,\text{s}^{-1}}$$
Substituting back, $\rho A_1 C_X = 1.21 \times 0.08081 \times 204.5 = 20.0\ \text{kg}\,\text{s}^{-1}$. This also validates the "air inlet velocity 200 m/s" quoted in Question 2 — the two questions describe the same machine.
Part (a)(iii) — evaluate the mean blade speed. Mid-height sits at the arithmetic mean diameter,
$$D_m = \tfrac{1}{2}\left(D_{\text{tip}} + D_{\text{root}}\right) = 0.2778\ \text{m}, \qquad U_m = \frac{\pi D_m N}{60}$$
$$\boxed{U_m = 240.0\ \text{m}\,\text{s}^{-1}}$$
Equivalently $U_m = 0.75\,U_{\text{tip}}$, since $D_m = 0.75\,D_{\text{tip}}$ when the hub ratio is one half.
Part (b) — close the inlet triangle. The guide vanes turn the flow $10^\circ$ from axial, so the inlet whirl and absolute velocity are
$$C_{Y1} = C_X\tan\alpha_1 = 204.5\tan 10^\circ = 36.06\ \text{m}\,\text{s}^{-1}, \qquad C_1 = \frac{C_X}{\cos\alpha_1} = 207.7\ \text{m}\,\text{s}^{-1}$$
Subtracting the whirl from the blade speed gives the relative components, $W_{Y1} = U_m - C_{Y1} = 240.0 - 36.06 = 203.9\ \text{m}\,\text{s}^{-1}$, hence
$$W_1 = \sqrt{C_X^{2} + W_{Y1}^{2}} = 288.8\ \text{m}\,\text{s}^{-1}, \qquad \beta_1 = \arctan\frac{W_{Y1}}{C_X} = 44.92^\circ$$
Part (b) concluded — close the outlet triangle. The rotor discharges at $\beta_2 = 30^\circ$ from axial with the axial velocity unchanged, so
$$W_{Y2} = C_X\tan\beta_2 = 118.1\ \text{m}\,\text{s}^{-1}, \qquad W_2 = \frac{C_X}{\cos\beta_2} = 236.2\ \text{m}\,\text{s}^{-1}$$
and the absolute outlet whirl follows from the same triangle closure, $C_{Y2} = U_m - W_{Y2} = 240.0 - 118.1 = 121.9\ \text{m}\,\text{s}^{-1}$, giving
$$C_2 = \sqrt{C_X^{2} + C_{Y2}^{2}} = 238.1\ \text{m}\,\text{s}^{-1}, \qquad \alpha_2 = \arctan\frac{C_{Y2}}{C_X} = 30.80^\circ$$
Both triangles satisfy $W_Y + C_Y = U$, which is the check a scale drawing is meant to provide.
Part (c) — apply the Euler work equation. The paper's own compressor relation on page 20 is $w = U(C_{Y2} - C_{Y1})$, so with the change of whirl $\Delta C_Y = 121.9 - 36.06 = 85.85\ \text{m}\,\text{s}^{-1}$,
$$w = 240.0 \times 85.85 = 20\,604\ \text{J}\,\text{kg}^{-1} = 20.60\ \text{kJ}\,\text{kg}^{-1}$$
$$P_1 = \dot{M}w = 20 \times 20\,604$$
$$\boxed{P_1 = 412.1\ \text{kW}}$$
The same number comes out of the rothalpy form $w = \tfrac{1}{2}\left[(C_2^{2} - C_1^{2}) + (W_1^{2} - W_2^{2})\right]$, and Question 2 will reach 410.2 kW from pure thermodynamics — agreement to 0.5 % confirms the angle convention used here.
