Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format: National Examination 07-Mec-A6-1 Fluid Machinery, December 2014 — closed book, three hours, 60 marks. Section A is calculative (Q1–Q5) and Section B is descriptive (Q6–Q8); the rubric asks for four questions of Section A plus two of Section B (six questions, each of equal value, 10 marks). All eight questions are solved in full as a study resource. General constants supplied with the paper: g = 9.81 m/s², patm = 100 kPa, pvapour = 2.34 kPa (20 °C), ρwater = 1000 kg/m³, ρair = 1.21 kg/m³ (15 °C), cp,air = 1.005, cv,air = 0.718 kJ/kg·K.
Reference texts. S. L. Dixon & C. A. Hall, Fluid Mechanics and Thermodynamics of Turbomachinery (7th ed.); R. A. Sabersky, A. J. Acosta, E. G. Hauptmann & E. M. Gates, Fluid Flow: A First Course in Fluid Mechanics (4th ed.) — source of the pump-selection charts Figs 15.11/15.12; H. Cohen, G. F. C. Rogers & H. I. H. Saravanamuttoo, Gas Turbine Theory; R. W. Fox, A. T. McDonald & P. J. Pritchard, Introduction to Fluid Mechanics.
Check: Questions 3 rely on chart reads from the examination attachments (Figs 15.11 & 15.12). Values read from those charts (φe, σc, η) are quoted to the precision the graphs allow and are flagged where used; a candidate would read the same values off the supplied plots. The induction-motor pole count is not stated in Q3, so a standard 4-pole (1800 rev/min synchronous) machine is assumed — the governing method is unaffected.
Approach. Get the peripheral speeds from ωr, the radial (meridional) velocities from continuity through the cylindrical flow areas, the outlet whirl from Euler's equation with no inlet swirl, then power and torque from Euler, and finally the blade angles from the two velocity triangles.
Radial (meridional) velocities. Continuity through the cylindrical area $2\pi r b$ gives$$V_{1R}=\frac{Q}{2\pi r_1 b_1}=\frac{0.25}{2\pi(0.10)(0.05)}=7.96\ \text{m/s},\qquad V_{2R}=\frac{Q}{2\pi r_2 b_2}=\frac{0.25}{2\pi(0.18)(0.03)}=7.37\ \text{m/s}.$$
Outlet whirl velocity (Euler, no inlet swirl). Pure radial inflow means $V_{1T}=0$, so Euler's pump equation $gH=V_{B2}V_{2T}-V_{B1}V_{1T}$ reduces to$$V_{2T}=\frac{gH}{V_{B2}}=\frac{9.81(40)}{32.4}=12.1\ \text{m/s}.$$$$\boxed{V_{2T}=12.1\ \text{m/s}}$$
Ideal power and shaft torque. With no losses the power is the Euler power $P=\rho Q\,V_{B2}V_{2T}=\rho gQH=1000(9.81)(0.25)(40)=98.1\ \text{kW}$, and the torque is the rate of change of angular momentum,$$\tau=\rho Q(r_2 V_{2T}-r_1 V_{1T})=1000(0.25)(0.18\times12.1)=545\ \text{N}\!\cdot\!\text{m}.$$$$\boxed{P=98.1\ \text{kW},\quad \tau=545\ \text{N}\!\cdot\!\text{m}}$$(check: $P=\tau\omega=545(180.1)=98.1\ \text{kW}$).
Blade angles from the velocity triangles. At inlet the flow is radial, so the relative velocity has tangential component $V_{B1}$ and radial component $V_{1R}$: $\tan\beta_1=V_{1R}/V_{B1}=7.96/18.0$, giving $\beta_1=23.8^\circ$. At outlet the relative tangential component is $V_{B2}-V_{2T}=32.4-12.1=20.3\ \text{m/s}$, so $\tan\beta_2=V_{2R}/(V_{B2}-V_{2T})=7.37/20.3$, giving $\beta_2=19.9^\circ$.$$\boxed{\beta_1=23.8^\circ,\quad \beta_2=19.9^\circ}$$
Fig. Q4a - Inlet velocity triangle: absolute inflow V1 is purely radial (no swirl), so the relative velocity W1 sets blade angle β₁ ≈ 23.8°.
