Question 5 of 6: Centrifugal Pump Performance – BEP and Affinity Scaling
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — December 2013 — 04-Agric-A4, Fluid Flow. Three-hour, open-book exam; a Casio or Sharp approved calculator is permitted. Format: four questions constitute a complete paper, with a choice between 1a/1b and between 4a/4b; all questions require calculation. Both alternatives are solved below for completeness.
Reference texts: White, Fluid Mechanics (7th/8th ed., McGraw-Hill) — open-channel hydraulics (hydraulic jump, gradually varied flow over a bump), pipe friction (Colebrook–White), rotating control volumes (sprinkler reaction), and turbomachinery (pump performance curves, affinity laws).
Question 4a: Centrifugal Pump Performance – BEP and Affinity Scaling (choose 4a or 4b — equal value)
Given. Measured $H$–$Q$–$P$ performance data for a centrifugal pump on water at 20°C (table below); water specific weight $\gamma=62.4\ \text{lbf/ft}^3$.
$Q$ (gal/min)
0
400
800
1200
1600
2000
2400
$H$ (ft)
123
115
108
101
93
81
62
$P$ (hp)
30
36
40
44
47
48
46
Find. (a) The best efficiency point (BEP) and maximum efficiency; (b) the most efficient flow rate, head, and brake horsepower after doubling the impeller diameter and increasing the speed by 50%.
Figure 4a — Pump head curve $H(Q)$ (solid) and computed efficiency curve $\eta(Q)$ (dashed) from the test data; the BEP is the peak of the efficiency curve.
Approach. Compute the hydraulic efficiency $\eta=\gamma Q H/(550P)$ at each tested point (with $Q$ converted to ft³/s), locate its peak with a local quadratic fit through the three highest points, then read $H$ and $P$ at that same flow from their own local quadratic fits. Rescale the BEP triplet to the new impeller diameter and speed using the pump affinity laws.
Efficiency at each test point. With $Q_{\text{cfs}}=Q_{\text{gpm}}(0.002228)$,
$$\eta = \frac{\gamma Q_{\text{cfs}} H}{550\,P}.$$
Evaluating at all seven points gives $\eta = 0,\ 32.3\%,\ 54.6\%,\ 69.6\%,\ 80.0\%,\ 85.3\%,\ 81.8\%$ for $Q=0$–$2400$ gpm — efficiency peaks between the 2000 and 2400 gal/min points.
Locate the BEP by local quadratic interpolation. Fit a parabola through the three highest-efficiency points ($Q=1600,2000,2400$ gpm; $\eta=80.0,85.3,81.8\%$) and take its vertex:
$$\boxed{Q_{\text{BEP}} \approx 2040\ \text{gal/min}}, \qquad \boxed{\eta_{\max} \approx 85.4\%}.$$
Head and power at the BEP. Evaluating the same-shaped local quadratics for $H(Q)$ and $P(Q)$ at $Q=2040$ gpm:
$$\boxed{H_{\text{BEP}} \approx 79.4\ \text{ft}}, \qquad \boxed{P_{\text{BEP}} \approx 47.9\ \text{hp}}$$
(cross-check: $\gamma Q_{\text{cfs}}H/550P = 85.4\%$, consistent with Step 2). That answers part (a).
Affinity-law scaling for part (b) — diameter doubled, speed ×1.5. For the same pump family (constant efficiency at the corresponding point),
$$\frac{Q_2}{Q_1}=\frac{N_2}{N_1}\left(\frac{D_2}{D_1}\right)^3,\quad \frac{H_2}{H_1}=\left(\frac{N_2}{N_1}\right)^2\left(\frac{D_2}{D_1}\right)^2,\quad \frac{P_2}{P_1}=\left(\frac{N_2}{N_1}\right)^3\left(\frac{D_2}{D_1}\right)^5,$$
with $N_2/N_1=1.5$ and $D_2/D_1=2$:
$$Q_2 = 2040(1.5)(2)^3 = 2040(12) = \boxed{Q_2 \approx 24{,}500\ \text{gal/min}},$$
$$H_2 = 79.4(1.5)^2(2)^2 = 79.4(9) = \boxed{H_2 \approx 715\ \text{ft}},$$
$$P_2 = 47.9(1.5)^3(2)^5 = 47.9(108) = \boxed{P_2 \approx 5170\ \text{hp}}.$$
