Exercise set

Environmental Water and Wastewater Systems Exercises

Worked environmental engineering exercises for water balances, wet-weather flow, pipe velocity, pump power, pollutant loading, equalization storage, residence time, outlet control, availability, and monitoring reconciliation.

These exercises practise first-pass calculations used in environmental water and wastewater systems. They connect water balances, wet-weather flow, pipe hydraulics, pumping, pollutant loading, equalization storage, residence time, outlet control, reliability, and monitoring reconciliation.

Assume simplified nominal values unless an exercise states otherwise. Real decisions require local standards, field data, hydraulic models, treatment-process limits, monitoring uncertainty, maintenance condition, permit requirements, and responsible engineering review.

How to Use These Exercises

For each problem:

  1. define the system boundary and operating case;
  2. separate average flow, peak flow, wet-weather flow, and abnormal events;
  3. use compatible time and volume units;
  4. connect hydraulic calculations to treatment, compliance, reliability, or safety consequences;
  5. identify the field data needed to validate the result.

The most common mistake is sizing from a single average flow. Water systems fail at boundaries: wet weather, pump outage, storage full, sensor drift, treatment peak load, pipe blockage, or downstream tailwater.

For each result, state whether it supports capacity sizing, wet-weather response, pump operation, treatment loading, storage control, outlet restriction, reliability improvement, or data-quality review. Hydraulic arithmetic should lead to an operating decision, not only a design number.

Exercise 1: Storage Water Balance

A storage basin receives inflow Q_{in}=0.18\ \text{m}^3/\text{s} and releases outflow Q_{out}=0.12\ \text{m}^3/\text{s}. Rainfall adds 0.015\ \text{m}^3/\text{s} across the basin surface, while evaporation and seepage remove 0.005\ \text{m}^3/\text{s}. The condition lasts for 2 hours.

Estimate the storage change.

Solution

Net storage rate:

\displaystyle \frac{dS}{dt}=Q_{in}-Q_{out}+P-E-I
\displaystyle \frac{dS}{dt}=0.18-0.12+0.015-0.005=0.070\ \text{m}^3/\text{s}

Time:

t=2(3600)=7200\ \text{s}

Storage change:

\Delta S=0.070(7200)=504\ \text{m}^3

Engineering Comment

The basin is filling. The next question is whether the available storage, outlet control, freeboard, and overflow route can safely handle the added volume.

Exercise 2: Wet-Weather Inflow and Infiltration

A wastewater catchment has dry-weather daily flow of 4200\ \text{m}^3/\text{day}. During a rainfall event, total daily flow rises to 7600\ \text{m}^3/\text{day}.

Estimate rainfall-derived inflow and infiltration, then express it as a percentage of wet-weather flow.

Solution

Rainfall-derived component:

Q_{RDI}=7600-4200=3400\ \text{m}^3/\text{day}

Percentage of wet-weather flow:

\displaystyle \frac{3400}{7600}\times100=44.7\%

Engineering Comment

Nearly half of the wet-weather flow is not normal sanitary flow under this simplified comparison. The system may need source tracing, manhole inspection, smoke testing, flow metering, private inflow reduction, or capacity review.

Exercise 3: Pipe Velocity Check

A pressure pipe carries Q=0.060\ \text{m}^3/\text{s} through diameter D=0.25\ \text{m}.

Find the cross-sectional area and average velocity.

Solution

Pipe area:

\displaystyle A=\frac{\pi D^2}{4}=\frac{\pi(0.25)^2}{4}=0.0491\ \text{m}^2

Velocity:

\displaystyle v=\frac{Q}{A}=\frac{0.060}{0.0491}=1.22\ \text{m/s}

Engineering Comment

The velocity is a screening value. Design review should also check head loss, minimum cleansing velocity, erosion, pressure class, surge risk, air release, fittings, and downstream hydraulic grade.

Exercise 4: Pump Input Power

A pump station must deliver Q=0.080\ \text{m}^3/\text{s} against total dynamic head H=22\ \text{m}. The combined pump and motor efficiency is \eta=0.68. Use \rho=1000\ \text{kg/m}^3 and g=9.81\ \text{m/s}^2.

Estimate input power.

Solution

Hydraulic input power relation:

\displaystyle P_{in}=\frac{\rho gQH}{\eta}
\displaystyle P_{in}=\frac{1000(9.81)(0.080)(22)}{0.68}=25{,}390\ \text{W}
P_{in}=25.4\ \text{kW}

Engineering Comment

Power is only one part of pump selection. The operating point should be checked against the system curve, minimum flow, pump cycling, wet-well volume, standby capacity, controls, surge, and backup power.

Exercise 5: Daily Pollutant Load

A wastewater treatment plant receives average flow Q=12{,}000\ \text{m}^3/\text{day}. Influent concentration of a pollutant is C=180\ \text{mg/L}.

Estimate daily pollutant load in \text{kg/day}.

Solution

Convert concentration:

180\ \text{mg/L}=0.180\ \text{kg/m}^3

Daily load:

M=QC
M=12{,}000(0.180)=2160\ \text{kg/day}

Engineering Comment

Treatment processes are sized by both hydraulic flow and mass loading. A plant can meet flow capacity and still fail if pollutant loading, temperature, solids, toxicity, or peak variation exceeds process assumptions.

Exercise 6: Equalization Storage

An equalization tank receives four one-hour inflow intervals:

0.18,\ 0.22,\ 0.10,\ 0.08\ \text{m}^3/\text{s}

The controlled outflow is 0.12\ \text{m}^3/\text{s}. Estimate the required active storage using cumulative positive storage from an initially empty operating point.

Solution

Storage increment:

\Delta S=(Q_{in}-Q_{out})\Delta t

with \Delta t=3600\ \text{s}:

\Delta S=(0.06,\ 0.10,\ -0.02,\ -0.04)(3600)
\Delta S=(216,\ 360,\ -72,\ -144)\ \text{m}^3

Cumulative storage:

S=(216,\ 576,\ 504,\ 360)\ \text{m}^3

Required active storage:

S_{req}=576\ \text{m}^3

Engineering Comment

The tank also needs operating volume, alarms, overflow handling, cleaning access, odor control, mixing, and emergency bypass review. Equalization volume is useful only when the control rule is practical.

Exercise 7: Nominal Residence Time

A contact tank has volume V=900\ \text{m}^3 and flow Q=0.15\ \text{m}^3/\text{s}.

Find the nominal residence time in hours and compare it with a required nominal value of 1.5 hours.

Solution

Residence time:

\displaystyle \tau=\frac{V}{Q}
\displaystyle \tau=\frac{900}{0.15}=6000\ \text{s}

Convert to hours:

\displaystyle \tau=\frac{6000}{3600}=1.67\ \text{h}

Since:

1.67>1.5

the nominal residence time exceeds the requirement.

Engineering Comment

Nominal residence time does not prove treatment performance. Short-circuiting, dead zones, baffle condition, temperature, mixing, disinfectant decay, and peak flow can reduce effective contact time.

Exercise 8: Orifice Outlet Flow

A basin outlet uses a circular orifice with diameter D=0.15\ \text{m}, discharge coefficient C_d=0.62, and head H=1.2\ \text{m}.

Estimate the outlet flow.

Solution

Area:

\displaystyle A=\frac{\pi D^2}{4}=\frac{\pi(0.15)^2}{4}=0.0177\ \text{m}^2

Orifice discharge:

Q=C_dA\sqrt{2gH}
Q=0.62(0.0177)\sqrt{2(9.81)(1.2)}
Q=0.053\ \text{m}^3/\text{s}

Engineering Comment

The actual outlet performance depends on debris, submergence, inlet geometry, tailwater, sediment, maintenance access, and whether the control remains unobstructed during the design event.

Exercise 9: Series Availability of a Pumping Function

A pumping function requires the pump, control panel, and power feed to work. Their estimated availabilities during the operating window are:

A_{pump}=0.97,\quad A_{control}=0.98,\quad A_{power}=0.99

Estimate simplified series availability.

Solution

Series availability:

A_{series}=A_{pump}A_{control}A_{power}
A_{series}=0.97(0.98)(0.99)=0.941
A_{series}=94.1\%

Engineering Comment

The pumping function is weaker than any single component because all three must work. Reliability review should consider standby pumps, backup power, alarms, spare parts, access during storms, and common-cause failures.

Exercise 10: Monitoring Balance Residual

A daily operations report shows inflow meter total 5200\ \text{m}^3/\text{day}, outflow meter total 4700\ \text{m}^3/\text{day}, and measured storage increase 200\ \text{m}^3/\text{day}.

Find the unexplained balance residual and express it as a percentage of inflow.

Solution

For a simple balance:

Residual=Q_{in}-Q_{out}-\Delta S
Residual=5200-4700-200=300\ \text{m}^3/\text{day}

Percentage of inflow:

\displaystyle \frac{300}{5200}\times100=5.8\%

Engineering Comment

The residual may indicate meter error, unmeasured overflow, leakage, infiltration, operational drawdown, timing mismatch, or incorrect storage measurement. A persistent residual should trigger data-quality review before the model is trusted.

Review Checklist

Before accepting an environmental water-system calculation, check:

  • whether the boundary and operating case are stated;
  • whether average, peak, wet-weather, and abnormal conditions are separated;
  • whether flow, volume, concentration, and time units are compatible;
  • whether storage has a defined operating rule and overflow path;
  • whether pump and pipe calculations include reliability and transient consequences;
  • whether pollutant load uses both flow and concentration;
  • whether residence time is validated against actual hydraulics, not only nominal volume;
  • whether telemetry, meter timing, storage readings, bypass logs, and rainfall records support the operating conclusion;
  • whether abnormal states such as pump outage, blocked outlet, peak load, tailwater, or power loss change the acceptance decision;
  • whether telemetry balances close well enough for the decision being made.

Good environmental water engineering connects calculations with monitoring, operations, maintenance, treatment performance, and compliance evidence.

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See also