Question 4 of 7

Problem 4 (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 (6 marks)

Direct runoff is the "excess" rainfall component that reaches the stream quickly — via overland flow and shallow interflow — during and shortly after a storm; it produces the sharp rising limb and peak of the hydrograph. Base flow is the sustained, slowly varying component of streamflow sourced from groundwater discharge into the channel; it persists between storms and is what keeps a stream flowing during dry weather. The recession curve is the falling limb of the hydrograph after the peak, and specifically its tail once direct runoff has essentially drained away, represents the depletion of bank/aquifer storage discharging back to the stream; it typically follows an exponential decay ($Q_t = Q_0 e^{-kt}$) and its shape (the recession constant $k$) is itself a useful indicator of the basin's groundwater storage characteristics. Hydrograph separation techniques use the recession curve, extrapolated backward under the storm peak, to split total streamflow into its base-flow and direct-runoff components.

(ii) Two main factors affecting hydrograph shape (6 marks)

1. Watershed physiographic characteristics (size, shape, slope and drainage density). A small, steep, compact watershed with a dense drainage network concentrates runoff quickly, producing a short time-to-peak, a high sharp peak and a rapid recession; a large, flat, elongated watershed spreads the runoff contribution over a much longer travel time, producing a lower, broader, more delayed peak. This directly affects hydrograph analysis because the unit hydrograph's time base and peak must be scaled (or an entirely different unit hydrograph derived) for basins of different physiographic character — a unit hydrograph is not transferable between dissimilar watersheds without adjustment.
2. Storm characteristics (rainfall intensity, duration and spatial/temporal distribution over the basin, including storm movement direction). A high-intensity, short-duration storm produces a sharply peaked hydrograph, while a low-intensity, long-duration storm produces a flatter, more prolonged response; a storm cell moving in the downstream direction of the drainage network increases the peak discharge (successive sub-basin contributions arrive in closer succession) compared with the same storm moving upstream. This matters for hydrograph analysis because the same total rainfall volume can produce very different peak flows and timing depending on how the storm is distributed, so a single design storm hyetograph must be chosen carefully (and storm movement considered) when deriving a design hydrograph.

(iii) Pump total dynamic head and brake horse power (8 marks)

Given.

Find. Total dynamic head (TDH) and brake horse power (BHP), with all losses other than pipe friction and static lift taken as negligible (as stated).

Approach. $\text{TDH} = \Delta\text{Elev} + H_f$ (Darcy–Weisbach friction loss added to the static lift), then the hydraulic (water) power $\rho g Q\,\text{TDH}$ is divided by pump efficiency to obtain the shaft (brake) power.

1. Velocity in the pipe. $A = \dfrac{\pi}{4}D^2 = \dfrac{\pi}{4}(0.295)^2 = 0.06835\ \text{m}^2$. $$V = \frac{Q}{A} = \frac{0.102}{0.06835} = 1.49\ \text{m/s}$$
2. Friction head loss (Darcy–Weisbach). $$H_f = f\,\frac{L}{D}\,\frac{V^2}{2g} = 0.014\times\frac{1500}{0.295}\times\frac{1.49^2}{2(9.81)} = 8.08\ \text{m}$$
3. Total dynamic head. With velocity head and minor losses stated negligible, $$\text{TDH} = \Delta\text{Elev} + H_f = 31 + 8.08 = \boxed{39.1\ \text{m}}$$
4. Brake horse power. Hydraulic power delivered to the water, $$P_{hyd} = \rho g Q\,\text{TDH} = 1000\times9.81\times0.102\times39.1 = 39{,}100\ \text{W} = 39.1\ \text{kW}$$ Dividing by pump efficiency for the required shaft input, $$\text{BHP} = \frac{P_{hyd}}{E_{pump}} = \frac{39.1}{0.70} = \boxed{55.9\ \text{kW} \approx 74.9\ \text{hp}}$$