Question 6 of 6: Pump System Operating Point at 880 r/min
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 4b: Pump System Operating Point at 880 r/min (choose 4a or 4b — equal value)
Given. The pump is the same backward-curved-blade centrifugal pump of Question 4a's test data (0–2400 gal/min table), now run at $N_2=880$ r/min through a 20 cm commercial-steel pipe.
Check: the source gives no test speed for the Question 4a performance table. This dataset is the well-known standard pump-curve set tested at $N_1=1170$ r/min (a common three-phase induction-motor synchronous speed for this class of problem); that assumption is used to scale the curve to $880$ r/min below.
Quantity
Value
Pump speed, $N_2$
880 r/min (test speed $N_1=1170$ r/min, assumed)
Pipe diameter, $D$
20 cm (commercial steel, $\epsilon=0.046$ mm)
Suction run (source to pump)
20 m, horizontal
Discharge run (pump to riser base)
12 m, horizontal
Riser (vertical rise to discharge level)
8 m
Source-tank submergence of intake
4 m below surface
Destination-tank submergence of outlet
3 m below surface
Check: fitting coefficients not given for the piping — standard values assumed: sharp-edged entrance $K=0.5$, two 90° flanged elbows (base and top of the riser) $K=0.3$ each, submerged exit $K=1.0$ ($\Sigma K = 2.1$), minor losses included alongside pipe friction.
Find. The delivered flow rate in ft³/min, and whether the pump runs efficiently at this duty.
Figure 4b — Pump system: static lift is the destination-surface elevation minus the source-surface elevation ($11\,\text{m}-4\,\text{m}=7\,\text{m}$); total pipe run is $20+12+8=40\,\text{m}$.
Approach. Fit the Question 4a data with a smooth $H_1(Q)$ curve, scale it to 880 r/min via the affinity laws, build the system head curve $H_{\text{sys}}(Q)=H_s+(fL/D+\Sigma K)V^2/2g$ with Colebrook friction, and solve the two simultaneously (iterating $f$ against $Re$) for the operating point.
Fit the 1170 r/min pump curve. A least-squares quadratic through all 7 data points (SI units, $Q$ in m³/s, $H$ in m) gives
$$H_1(Q) = 36.94 - 44.00\,Q - 467.2\,Q^2 \quad (R^2 > 0.999).$$
Scale to $N_2=880$ r/min. Since $H\propto N^2$ at fixed $Q/N$, a quadratic $H_1=a_1+b_1Q+c_1Q^2$ scales as $H_2(Q)=a_1(N_2/N_1)^2+b_1(N_2/N_1)Q+c_1Q^2$. With $N_2/N_1=880/1170=0.7521$:
$$H_2(Q) = 20.90 - 33.09\,Q - 467.2\,Q^2 \quad (\text{m},\ Q\ \text{in m}^3/\text{s}).$$
(Shutoff head at 880 r/min: $20.9\,\text{m}\approx69\,\text{ft}$, versus $123\,\text{ft}$ at 1170 r/min — consistent with $(N_2/N_1)^2=0.566$.)
Static lift and system geometry. Taking the suction/discharge pipe centerline as datum, the source surface sits at $+4\,$m and the destination surface at $+(8+3)=+11\,$m, so
$$H_s = 11-4 = \boxed{7.0\ \text{m}}.$$
Total pipe length $L=20+12+8=40\,$m, $D=0.20\,$m, $\epsilon/D=0.046/200=2.3\times10^{-4}$, $\Sigma K = 0.5+2(0.3)+1.0=2.1$.
System head curve.
$$H_{\text{sys}}(Q) = H_s + \left(f\frac{L}{D}+\Sigma K\right)\frac{V^2}{2g}, \qquad V=\frac{Q}{A},\ A=\frac{\pi}{4}(0.20)^2=0.03142\ \text{m}^2.$$
Solve $H_2(Q)=H_{\text{sys}}(Q)$ iteratively. Guess $Q$, get $Re=\rho VD/\mu$, solve Colebrook for $f$, solve the resulting quadratic in $Q$, repeat until $f$ stops changing. This converges in a couple of passes ($f\approx0.0153$, fully turbulent, changes little with $Q$ here) to
$$\boxed{Q \approx 0.1169\ \text{m}^3/\text{s}} \;\Rightarrow\; \boxed{Q \approx 248\ \text{ft}^3/\text{min}}\ (\approx 1854\ \text{US gal/min}),$$
with $V=3.72\ \text{m/s}$, $Re\approx7.4\times10^5$, $f=0.0153$, and matched head $H_2=H_{\text{sys}}\approx10.6\ \text{m}\ (34.9\ \text{ft})$. Brake power at this point scales similarly to $\approx 20\ \text{hp}$.
Is this an efficient application? The operating point corresponds to an equivalent flow of $Q/(N_2/N_1)\approx2460$ gal/min at the 1170 r/min test speed — just past the BEP found in Question 4a ($\approx2040$ gal/min, $\eta_{\max}\approx85\%$). Reading the same efficiency curve there gives $\eta\approx83\%$ at the operating point: still close to peak, but the system pulls the pump about 20% past its best-efficiency flow.
