Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format: National Examination 98-Mar-B5 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.
the same misfiling pattern already documented for other 98-Mar-coded papers in this discipline. It is solved here exactly as printed. Separately, Questions 3 relies 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. 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.
Question 3: Pump Application and Cavitation (10 marks)
Approach. Fix the running speed from the 60 Hz motor (4-pole, 3% slip), compute the dimensionless specific speed, read pump type, head coefficient φe and cavitation parameter σc off Figs 15.11/15.12, then size the impeller, evaluate NPSH and the suction-lift limit and finish with the hydraulic→shaft→electrical power chain.
Running speed and specific speed. A 4-pole 60 Hz motor has synchronous speed 1800 rev/min; at 3% slip $N=1800(0.97)=1746\ \text{rev/min}$, i.e. $\omega_e=182.8\ \text{rad/s}$. The dimensionless (SI) specific speed is$$(N_s)_{SI}=\frac{\omega_e\sqrt{Q}}{(gH)^{3/4}}=\frac{182.8\sqrt{0.085}}{(9.81\times30)^{3/4}}=0.75.$$$$\boxed{(N_s)_{SI}=0.75}$$
Pump type (Fig 15.11). At $(N_s)_{SI}\approx0.75$ (about 2000 on the gpm scale) the operating point lies in the radial-flow band, so a conventional radial (centrifugal) impeller with backward-curved vanes is appropriate — sketched above.
Impeller diameter. From Fig 15.11 the head (peripheral-velocity) factor at this specific speed is $\phi_e\approx1.0$. With $\phi_e=V_{B2}/\sqrt{2gH}$ the blade-tip speed is $V_{B2}=\phi_e\sqrt{2gH}=1.0\sqrt{2(9.81)(30)}=24.3\ \text{m/s}$, and $V_{B2}=\pi D N/60$ gives$$D=\frac{60\,V_{B2}}{\pi N}=\frac{60(24.3)}{\pi(1746)}=0.265\ \text{m}.$$$$\boxed{D\approx265\ \text{mm}}$$
Critical cavitation parameter (Fig 15.12). At $(N_s)_{SI}=0.75$ the chart gives $\sigma_c\approx0.30$.
Desired NPSH. With $\sigma_c=\text{NPSH}/H$,$$\text{NPSH}=\sigma_c H=0.30(30)=9.0\ \text{m}.$$$$\boxed{\text{NPSH}_{req}=9.0\ \text{m}}$$
Maximum pump elevation above supply. The available NPSH is $\text{NPSH}_{av}=\dfrac{p_{atm}-p_v}{\rho g}-\Delta z-h_L$. Setting it equal to the required NPSH and solving for the allowable static lift,$$\Delta z_{max}=\frac{p_{atm}-p_v}{\rho g}-h_L-\text{NPSH}=\frac{98\,000}{9810}-1.0-9.0=-0.01\ \text{m}.$$$$\boxed{\Delta z_{max}\approx0}$$i.e. the pump must sit essentially at (or just below) the reservoir supply level — no suction lift is available at this head.
Pump efficiency (Fig 15.11). The efficiency curve at $(N_s)_{SI}=0.75$ reads $\eta_p\approx0.85$ (85%).
Electric power consumption. The hydraulic (water) power is $P_w=\rho gQH=1000(9.81)(0.085)(30)=25.0\ \text{kW}$; dividing by pump efficiency gives the shaft power $P_s=25.0/0.85=29.4\ \text{kW}$. The motor efficiency combines slip and electrical losses, $\eta_m=(1-0.03)(1-0.04)=0.931$, so$$P_{elec}=\frac{P_s}{\eta_m}=\frac{29.4}{0.931}=31.6\ \text{kW}.$$$$\boxed{P_{elec}=31.6\ \text{kW}}$$
Check: φe ≈ 1.0, σc ≈ 0.30 and ηp ≈ 0.85 are read from the supplied charts and carry graph-reading precision (±a few %). The maximum lift comes out to essentially zero, which is the physically instructive result: at 30 m head the ≈10 m of atmospheric head is almost entirely consumed by the required NPSH and inlet losses, so the pump is best located at or below the supply level.
