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18-Env-A2 Hydrology and Municipal Hydraulics Engineering · May 2018

Question 1 of 7

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Exams — May 2018 — 04-Env-A2 / Hydrology and Municipal Hydraulics Engineering. 3 hours duration; closed book with an 8.5×11 in double-sided aid sheet; Casio or Sharp approved calculator only. Any five questions constitute a complete paper (only the first five answers in the work book are marked); all seven Problems are solved below for completeness. Each question is worth 20 marks.

Reference texts. Davis & Cornwell, Introduction to Environmental Engineering (6th ed.); Linsley, Kohler & Paulhus, Hydrology for Engineers (3rd ed.); Chow, Open-Channel Hydraulics; Walski et al., Advanced Water Distribution Modeling and Management; Metcalf & Eddy, Wastewater Engineering: Treatment and Resource Recovery (5th ed.); Guidelines for Canadian Drinking Water Quality (Health Canada); Canadian Council of Ministers of the Environment (CCME) water-quality guidelines.

Problem 1 (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) Water balance equation for the conceptual land-flow model (8 marks)

For any storage element (or the whole land phase) over a time interval, continuity requires that inflow minus outflow equals the change in storage. Written for the whole pervious-land system shown in the figure:

$$P = E_a + Q_s + Q_i + Q_g + \Delta S$$

where $P$ is precipitation plus snowmelt input, $E_a$ is actual evapotranspiration (the sum of the dashed-arrow losses from interception, upper-soil, lower-soil and groundwater storage), $Q_s$ is overland flow, $Q_i$ is interflow, $Q_g$ is the net outflow to deep/inactive groundwater, and $\Delta S$ is the net change in ALL storages (interception, upper soil zone, lower soil zone and active groundwater) over the interval.

How it is integrated into the model. The schematic is a cascade of storage "buckets" connected by flow paths, and the water-balance equation above is applied locally at every node as well as globally over the whole system. Snowmelt and precipitation first fill Interception Storage; storage in excess of interception capacity reaches the ground and either runs off immediately as Overland Flow or enters the soil at the Infiltration/Inflow junction. From there the model routes water vertically through Upper Soil Zone Storage (which itself spills to Interflow once its capacity is exceeded, and passes Delayed Infiltration downward), through Lower Soil Zone Storage, and finally into Active Groundwater Storage, part of which is lost to Deep/Inactive Groundwater and the rest of which eventually reappears as baseflow. Each of the four storages also loses water directly to Actual ET (the dashed arrows), so the equation above is really four coupled continuity equations — one per storage — solved in sequence at every time step; their sum, checked against total inflow $P$, is exactly the water-balance equation and is what lets the conceptual model be calibrated and mass-balance-checked against an observed streamflow record.

[Figure not reproduced: Fig. 1 — Conceptual storage-cascade model for pervious land flow (as printed on the exam), redrawn to show the routing paths referenced in the water-balance explanation. See the official exam paper.]

(ii) PVC pipe flow — velocity, Reynolds number, friction loss (6 marks)

Given.

QuantityValue
Pipe length, $L$100 m
Pipe diameter, $d$200 mm = 0.200 m
Full flow rate, $Q$50 L/s = 0.050 m³/s
Pipe materialPVC, roughness $\varepsilon\approx1.5\times10^{-6}\ \text{m}$
Kinematic viscosity, $\nu$ (water, 20 °C)$1.004\times10^{-6}\ \text{m}^2/\text{s}$

Find. Average velocity $V$; Reynolds number $Re$ and flow regime; friction head loss $H_f$.

Approach. Compute $V$ from continuity, classify the flow with $Re$, then solve the Colebrook–White equation for the Darcy friction factor $f$ and apply the Darcy–Weisbach equation for $H_f$.

  1. Average velocity. $$A = \frac{\pi d^2}{4} = \frac{\pi(0.200)^2}{4} = 0.03142\ \text{m}^2 \qquad V = \frac{Q}{A} = \frac{0.050}{0.03142} = \boxed{1.59\ \text{m/s}}$$
  2. Reynolds number and flow regime. $$Re = \frac{Vd}{\nu} = \frac{(1.59)(0.200)}{1.004\times10^{-6}} = \boxed{3.17\times10^{5}}$$ Since $Re \gt 4000$, the flow is turbulent.
  3. Friction factor and head loss. With relative roughness $\varepsilon/d = 7.5\times10^{-6}$, solving the Colebrook–White equation iteratively gives $f=0.0144$. Applying Darcy–Weisbach, $$H_f = f\,\frac{L}{d}\,\frac{V^2}{2g} = (0.0144)\left(\frac{100}{0.200}\right)\frac{(1.59)^2}{2(9.81)} = \boxed{0.929\ \text{m}}$$
QuantityValue
Average velocity, $V$1.59 m/s
Reynolds number, $Re$3.17×10&sup5; (turbulent)
Friction factor, $f$0.0144
Friction head loss, $H_f$0.929 m

(iii) Functions of a low-lift pumping station (6 marks)

A low-lift pumping station is the first pump station in a water treatment plant, lifting raw water from the intake (river, lake or reservoir, often at a level well below the plant) into the head of the treatment train. Three of its main functions are: (1) overcoming the initial static lift and intake/screen losses so that raw water reaches the coagulation/flocculation and sedimentation basins with enough elevation to flow the rest of the way through the plant by gravity; (2) providing a controlled, reasonably constant flow rate into the flocculation basins despite a fluctuating source water level (seasonal river/lake stage changes) or varying intake screen head loss, which is important because flocculation and sedimentation both depend on a stable detention time to work effectively; and (3) providing flow-pacing and redundant capacity — low-lift stations are normally built with multiple pumps (duplex/triplex) so flow can be matched to plant demand and so that one unit can be taken out of service for maintenance without interrupting the raw-water supply to the plant.

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