Question 3 of 7: Centrifugal Pump Characteristics and a Firefighting System

Question 3: Centrifugal Pump Characteristics and a Firefighting System (20 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. A single pump's Q–H–η curve at 1000 rpm (5 points), a 7.35 kW drive motor, and – for the firefighting duty – two such pumps in series feeding a 200 m × 100 mm pipe (friction factor 0.02) with an 8 m static lift and a 50 mm nozzle ($C_v=0.97$) at the end.

Find. The specific speed at best efficiency, the maximum discharge the 7.35 kW motor can sustain, and – for two pumps in series against the firefighting pipeline – the operating point, the power required and a judgement on suitability.

Approach. Read the specific speed from the best-efficiency point on the curve; find the discharge at which the hydraulic shaft-power demand $\rho gQH/\eta$ exactly equals the motor's rated power (that is the maximum sustainable discharge); then build the system-head curve for two pumps in series against the pipeline + nozzle and intersect it with twice the single-pump curve to get the operating point, and evaluate whether that point makes efficient, practical sense for a fire pump.

1. Specific speed at BEP. The efficiency peaks at Q=20 l/s, H=22.5 m ($\eta=73.5\%$). Using the metric form $N_s=N\sqrt{Q}/H^{0.75}$ (N in rpm, Q in m³/s, H in m): $$N_s=\frac{1000\sqrt{0.020}}{22.5^{0.75}}=\frac{1000(0.1414)}{10.33}=\boxed{13.7}$$ – a low specific speed, characteristic of a small radial-flow (high-head-for-its-flow) centrifugal pump.
2. Maximum discharge on a 7.35 kW motor. Required shaft power at each tabulated point is $P=\rho gQH/\eta$; tabulating gives 3.73, 5.05, 6.01, 7.47 and 8.19 kW at Q=10…27 l/s – the required power passes 7.35 kW between Q=20 and Q=25 l/s. Linear interpolation on the bracketing points: $$Q_{max}=20+\frac{7350-6006}{7470-6006}(25-20)=\boxed{24.6\ \text{l/s}}$$ Beyond this the pump would demand more than the motor can deliver, so this is the ceiling on discharge.
3. System head curve for the firefighting line. $A_{pipe}=\tfrac{\pi}{4}(0.100)^2=7.854\times10^{-3}\ \text{m}^2$, $A_{nozzle}=\tfrac{\pi}{4}(0.050)^2=1.963\times10^{-3}\ \text{m}^2$. Head required $=$ static lift $+$ Darcy pipe friction $+$ the velocity head needed to produce the nozzle jet allowing for its velocity coefficient: $$H_{sys}(Q)=H_s+f\frac{L}{D}\frac{V_{pipe}^2}{2g}+\frac{V_{nozzle}^2}{2gC_v^2}$$ with $V_{pipe}=Q/A_{pipe}$, $V_{nozzle}=Q/A_{nozzle}$ (same Q, continuity).
4. Operating point – two pumps in series. Two pumps in series at the same Q double the head: $H_{2pumps}(Q)=2H_{single}(Q)$ (single-pump H read by linear interpolation on the table). Intersecting $2H_{single}(Q)$ with $H_{sys}(Q)$ (numerically, bisection) gives $$\boxed{Q_{op}\approx 26.5\ \text{l/s}, \quad H_{op}\approx 41.2\ \text{m}}$$ at which $h_f=23.3$ m, $h_{nozzle}=9.9$ m, $V_{pipe}=3.38$ m/s, $V_{nozzle}=13.5$ m/s, and each pump runs at $\eta\approx66.8\%$ (below its 73.5% BEP, since the series-pump system pushes flow above the curve's sweet spot).
5. Power required. Each of the two pumps handles the full Q at its own single-pump head, so total shaft power is $$P_{total}=2\times\frac{\rho g Q_{op} H_{single}(Q_{op})}{\eta_{op}}=\boxed{16.0\ \text{kW}}$$

Is this a good choice? The series pair is well within its safe operating range (26.5 l/s sits inside the tested 10–27 l/s curve, so no extrapolation risk) and delivers a workable 13.5 m/s nozzle jet, but it operates about 7 percentage points below peak efficiency and close to the top end of the curve where efficiency is already falling – a pump sized with slightly more head margin (or a smaller nozzle to move the system curve's operating point back toward Q=20 l/s) would both save power and extend bearing/seal life. As specified it is workable but not optimal.