Question 1 of 4: Centrifugal pump — best efficiency point and similarity scaling

Question 1: Centrifugal pump — best efficiency point and similarity scaling (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. The seven-point test table above (water, 20°C, γ_water≈62.4 lbf/ft³). No original speed N₀ or diameter D₀ is stated — part (b) only needs the ratios D₂/D₁ = 2 and N₂/N₁ = 1.5.

Find. (a) the flow rate, head and brake horsepower at best efficiency, and the maximum efficiency; (b) the same three quantities for a geometrically similar pump twice the diameter, running 50% faster.

Approach. Compute the hydraulic efficiency at each test point, fit a parabola through the three points bracketing the observed peak to refine the BEP, then scale that operating point to the larger, faster homologous pump using the dimensionless pump-similarity coefficients.

1. Compute the hydraulic efficiency at every test point. Efficiency is the ratio of fluid (water) power delivered to shaft power supplied, $\eta = \dfrac{\gamma Q H}{P}$, with $Q$ converted from gal/min to ft$^3$/s ($1$ gal/min $= 2.228\times10^{-3}$ ft$^3$/s) and $P$ from hp to ft·lbf/s ($\times 550$). For example at $Q=2000$ gal/min ($Q=4.456$ ft$^3$/s, $H=81$ ft, $P=48$ hp): $$\eta = \frac{(62.4)(4.456)(81)}{(48)(550)} = \frac{22\,523}{26\,400} = 0.853$$ Repeating for every column gives:

   Q, gal/min	400	800	1200	1600	2000	2400
   η, %	32.3	54.6	69.6	80.0	85.3	81.8

   The tabulated maximum is at $Q=2000$ gal/min, bracketed by 1600 and 2400 gal/min — the true (continuous) peak lies between the 2000 and 2400 gal/min points, since efficiency is still rising steeply from 1600 to 2000 but has only started to fall by 2400.
2. Locate the best efficiency point by a parabolic fit. Fitting a parabola through the three bracketing points $(1600,\,80.03\%)$, $(2000,\,85.31\%)$, $(2400,\,81.77\%)$ and solving for its vertex gives the refined estimate $$\boxed{Q^{*} \approx 2039\ \text{gal/min}, \qquad \eta_{max} \approx 85.3\%}$$ Interpolating $H$ and $P$ with the same three-point parabola at $Q=Q^{*}$ gives $H^{*}\approx 79\ \text{ft}$ and $P^{*}\approx 48\ \text{hp}$ (consistent with the raw $Q=2000$ row, since the vertex sits only 39 gal/min away).
3. Scale the BEP to the larger, faster pump using similarity laws. Two centrifugal pumps of the same family (geometrically similar impellers) operate at matching (dynamically similar) points — in particular the same best-efficiency point — when their dimensionless coefficients match: $$C_Q=\frac{Q}{ND^3}, \qquad C_H=\frac{gH}{N^2D^2}, \qquad C_P=\frac{P}{\rho N^3D^5}$$ Holding these constant between the tested pump (subscript 1) and the modified pump (subscript 2) gives the similarity ratios $$\frac{Q_2}{Q_1}=\frac{N_2}{N_1}\left(\frac{D_2}{D_1}\right)^{3}, \qquad \frac{H_2}{H_1}=\left(\frac{N_2}{N_1}\right)^{2}\left(\frac{D_2}{D_1}\right)^{2}, \qquad \frac{P_2}{P_1}=\left(\frac{N_2}{N_1}\right)^{3}\left(\frac{D_2}{D_1}\right)^{5}$$ With $D_2/D_1=2$ and $N_2/N_1=1.5$: the multipliers are $1.5\times2^3=12$ for $Q$, $1.5^2\times2^2=9$ for $H$, and $1.5^3\times2^5=108$ for $P$. Applying them to the BEP found in Step 2, $$\boxed{Q_2 \approx 2039 \times 12 \approx 24\,500\ \text{gal/min}, \quad H_2 \approx 79\times 9 \approx 715\ \text{ft}, \quad P_2 \approx 48\times108 \approx 5180\ \text{hp}}$$ The power multiplier (108) dominates — doubling the diameter alone would multiply power by $2^5=32$, and the added 50% speed increase compounds that another $1.5^3\approx3.4$×, which is why brake horsepower grows so much faster than flow rate under combined scaling.