Question 3 of 8: Pump Application and Cavitation

Question 3: Pump Application and Cavitation (10 marks)

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.

Given. Q = 85 L/s = 0.085 m³/s; H = 30 m; 60 Hz motor, 3% slip, 4% electrical losses; inlet head loss h_L = 1.0 m; vapour pressure p_v = 2 kPa; p_atm = 100 kPa; ρ = 1000 kg/m³.

Find. N_s, pump type, impeller diameter, σ_c, NPSH, maximum suction lift, efficiency and electrical power.

Approach. Fix the running speed from the 60 Hz motor (4-pole, 3% slip), compute the dimensionless specific speed, read pump type, head coefficient φ_e and cavitation parameter σ_c 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.

1. 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$$(N_s)_{SI}=\frac{\omega_e\sqrt{Q}}{(gH)^{3/4}}=\frac{182.8\sqrt{0.085}}{(9.81\times30)^{3/4}}=0.75.$$$$\boxed{(N_s)_{SI}=0.75}$$
2. 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.
3. Impeller diameter. From Fig 15.11 the head (peripheral-velocity) factor at this specific speed is $\phi_e\approx1.09$ (the $\phi_e$ curve crosses 1.0 at $(N_s)_{SI}\approx0.56$ and is climbing). With $\phi_e=V_{B2}/\sqrt{2gH}$ the blade-tip speed is $V_{B2}=\phi_e\sqrt{2gH}=1.09\sqrt{2(9.81)(30)}=26.4\ \text{m/s}$, and $V_{B2}=\pi D N/60$ gives$$D=\frac{60\,V_{B2}}{\pi N}=\frac{60(26.4)}{\pi(1746)}=0.289\ \text{m}.$$$$\boxed{D\approx289\ \text{mm}}$$
4. Critical cavitation parameter (Fig 15.12). Fig 15.12 plots $\sigma_c$ against specific speed on linear axes; the curve passes through $\sigma_c\approx0.6$ at $(N_s)_{SI}=2$ and $\approx1.0$ at $(N_s)_{SI}=3$, and bends toward the origin below $(N_s)_{SI}\approx1.5$. At $(N_s)_{SI}=0.75$ it reads$$\sigma_c\approx0.15.$$
5. Desired NPSH. With $\sigma_c=\text{NPSH}/H$,$$\text{NPSH}=\sigma_c H=0.15(30)=4.5\ \text{m}.$$$$\boxed{\text{NPSH}_{req}=4.5\ \text{m}}$$
6. 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,$$\Delta z_{max}=\frac{p_{atm}-p_v}{\rho g}-h_L-\text{NPSH}=\frac{98\,000}{9810}-1.0-4.5=+4.5\ \text{m}.$$$$\boxed{\Delta z_{max}\approx+4.5\ \text{m}}$$i.e. the pump may be set up to about 4.5 m above the supply water level before the suction pressure falls to the cavitation limit. Of the 9.99 m of atmospheric head available, 1.0 m is spent on inlet friction and 4.5 m must be held back as NPSH, leaving 4.5 m of static lift.
7. Pump efficiency (Fig 15.11). The optimum-efficiency curve at $(N_s)_{SI}=0.75$ reads $\eta_p\approx0.92$ (92%) — the chart is already close to its 93% crest, which it reaches near $(N_s)_{SI}\approx0.9$.
8. 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 $P_s=25.0/0.92=27.2\ \text{kW}$. The motor efficiency combines slip and electrical losses, $\eta_m=(1-0.03)(1-0.04)=0.931$, so$$P_{elec}=\frac{P_s}{\eta_m}=\frac{27.2}{0.931}=29.2\ \text{kW}.$$$$\boxed{P_{elec}=29.2\ \text{kW}}$$