Question 4 of 7: Hydrograph Analysis and Pump Impeller Affinity Laws

Question 4: Hydrograph Analysis and Pump Impeller Affinity Laws (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.

(i) Recession Curve, Base Flow, and Direct Runoff

Base flow is the sustained component of streamflow contributed by groundwater discharge and delayed subsurface drainage; it is present between storms and forms the "floor" on which any storm hydrograph is superimposed. Direct runoff is the portion of the measured hydrograph directly attributable to a specific rainfall event (surface runoff plus, in some definitions, fast interflow); it is obtained by subtracting an estimated base-flow line from the total hydrograph, and it is this separated volume that is normally used for unit-hydrograph analysis of the storm. The recession curve is the declining tail of the hydrograph after direct runoff has passed, representing the basin's gradual depletion of its groundwater/channel storage; it is commonly modelled as an exponential decay $Q_t = Q_0 e^{-kt}$, where the recession constant $k$ is a characteristic property of the basin's storage and drainage efficiency.

(ii) Factors Affecting Hydrograph Shape

(1) Watershed characteristics — size, shape, slope, drainage density and land use/imperviousness. A small, steep, highly impervious (urbanized) watershed with an efficient drainage network produces a "flashy" hydrograph: short time-to-peak, high peak discharge, and steep rising and falling limbs, because impervious surfaces and engineered conveyance shorten travel time and minimize infiltration/depression-storage losses. This directly impacts hydrograph analysis because a unit hydrograph derived under one land-use condition cannot be transferred to a different (e.g. post-development) condition without an explicit correction for the change in basin lag and peak.

(2) Storm characteristics — rainfall intensity, duration and spatial distribution relative to the watershed. A short, intense storm concentrated near the watershed outlet produces a sharper, higher, earlier peak than the same total rainfall volume spread over a longer duration or centred near the watershed's most distant point. This impacts analysis because unit-hydrograph theory strictly assumes a rainfall duration close to the standard duration the unit hydrograph was derived for; storms of a substantially different duration require S-curve conversion before the unit hydrograph can be validly applied.

(iii) Pump Impeller Diameter Change — Affinity Laws

Given. A single pump casing with an interchangeable impeller; characteristic curve family (Total head vs. Capacity, with nested efficiency contours and constant brake-power lines) supplied for impeller diameters 90–150 mm:

Find. The new (130 mm) optimum capacity, head, brake horsepower and efficiency, and the percent capacity improvement.

Approach. Read the 95 mm impeller's best operating point off the chart, then scale it to the 130 mm impeller with the pump affinity laws for a geometrically similar diameter change at constant rotational speed; cross-check the scaled point against the actual 130 mm curve on the chart.

1. Compute the brake power at the original operating point. Hydraulic (water) power then shaft (brake) power: $$P_{hyd,1} = \rho g Q_1 H_1 = (1000)(9.81)(0.0084)(7.0) = 577\ \text{W}, \qquad BHP_1 = \frac{P_{hyd,1}}{\eta_1} = \frac{0.577}{0.63} = \boxed{0.92\ \text{kW}}.$$
2. Apply the affinity laws. For the same casing and speed, a change in impeller diameter scales capacity linearly, head quadratically and power cubically, with efficiency approximately unchanged between similar operating points: $$\frac{Q_2}{Q_1} = \frac{D_2}{D_1}, \qquad \frac{H_2}{H_1} = \left(\frac{D_2}{D_1}\right)^2, \qquad \frac{BHP_2}{BHP_1} = \left(\frac{D_2}{D_1}\right)^3, \qquad \eta_2 \approx \eta_1.$$ With $D_2/D_1 = 130/95 = 1.368$: $$Q_2 = (0.0084)(1.368) = \boxed{0.0115\ \text{m}^3/\text{s}}, \qquad H_2 = (7.0)(1.368)^2 = \boxed{13.1\ \text{m}}.$$
3. Scale the brake power and read the new efficiency. $$BHP_2 = (0.92)(1.368)^3 = \boxed{2.35\ \text{kW}}, \qquad \eta_2 \approx 63\%.$$ Cross-checking on the chart: the actual 130 mm impeller curve passes through $H\approx13$ m at $Q\approx0.0115$ m³/s — essentially the same point predicted by the affinity laws, confirming the two readings are mutually consistent.
4. Percent capacity improvement. $$\%\,\text{improvement} = \frac{Q_2-Q_1}{Q_1}\times100\% = \left(\frac{D_2}{D_1}-1\right)\times100\% = \boxed{36.8\%}.$$