Water Treatment

Environmental Engineering · FE Reference Handbook section

Environmental Engineering

0 formulas

10 exam-style examples

~45 min

All Environmental Engineering lectures

Learning objectives

What you must be able to do before leaving this section.

This chapter section covers Water Treatment within Environmental Engineering. Read it the way you would read a textbook chapter: the theory first so the relations mean something, then every equation with its use and its trap, then 10 fully worked examples with the arithmetic shown line by line, and finally a self-check you should be able to answer without notes.

Explain, in your own words, what water treatment describes physically and when it applies.
State every one of the 0 relations the handbook lists here and name each symbol with its unit.
Select the correct relation from the wording of an exam stem within 20 seconds.
Carry a complete solution from givens to a "most nearly" answer with the correct unit.
Recognise the distractors generated by the unit trap: mg/L × MGD × 8.34 = lb/day is the single most used conversion.

Lecture

Why this section exists. Water Treatment is the part of Environmental Engineering that lets you connect a treatment unit or receiving water body to a number you can defend. Before any equation is useful you must be able to picture the physical situation it describes; the schematic below is that picture.

How the theory is built. The handbook prints results, not derivations. Each relation in this section comes from one governing principle applied to the idealised system: state the principle, impose the stated assumptions, and the printed equation follows. Knowing which assumption each relation rests on is what lets you reject a wrong answer choice in seconds.

How it is examined. Items from this page are written as a mass balance across one reactor or one unit process. Roughly two thirds are direct substitution, one third require one intermediate quantity from a neighbouring relation, and a small number are conceptual — testing whether you know the assumption, not the arithmetic.

The habit that earns the points. Unit discipline. mg/L × MGD × 8.34 = lb/day is the single most used conversion. Every relation below is dimensionally consistent only when that rule is honoured, and the distractor set is deliberately built from candidates who ignored it. Write the unit next to every number you substitute, every time.

How to study this page. Read the theory, then cover the formula cards and try to reproduce each relation from its description. Then work the examples with the solution hidden, revealing one line at a time. Finish with the self-check questions; if you cannot answer one, return to the matching formula card.

Aeration basin at a wastewater treatment plant with churning aerated water and walkways. — Photo 1. Where this shows up in practice: water treatment.
Capstone Studio instructional photograph

Environmental Engineering — Water Treatment: reference schematic for orienting the symbols used in this section.

Theory, developed

Read this before the equations — it is what makes them memorable.

The physical situation

Every item from this section describes a treatment unit or receiving water body. Sketch it before you compute — a labelled sketch with the givens on it converts a wordy stem into a solvable problem and exposes the quantity the examiner left out on purpose.

The governing principle

The 0 relations on this page are consequences of one principle applied to that idealised system. Identify which quantity is conserved, balanced, or defined, and the correct equation follows without memorisation.

Assumptions and limits of validity

Each printed relation carries silent assumptions — linearity, steady state, uniformity, small deformation, or standard conditions, depending on the subject. Conceptual exam items are written by violating exactly one of these, so read the sentence above the equation as carefully as the equation itself.

Solution procedure you should automate

1) Read the last sentence of the stem to identify the requested quantity. 2) Locate the relation on this page whose left-hand side is that quantity. 3) Tabulate the givens with units and mark the missing symbol. 4) If a symbol is missing, find the one relation that produces it. 5) Rearrange symbolically, substitute once, evaluate, and round only at the end.

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

Clarification following coagulation and flocculation:
Alum coagulation 350−550 14−22 4−8 12−16
Ferric coagulation 550−700 22−28 4−8 12−16
Upflow clarifiers
Groundwater 1,500−2,200 61−90 1
Surface water 1,000−1,500 41−61 4

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

This section is conceptual; there are no equations to memorise.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1

Remaining BOD after five days

A wastewater has BOD_u = 320 mg/L with k = 0.23 /day (base e). What is BOD₅ and the remaining oxygen demand at day 5?

Given

BOD_u = 320 mg/L
k = 0.23 /day
t = 5 days

Find

BOD₅ and the remaining demand

Start with the thinking

BOD₅ is the amount exerted, not what remains.
Exponential decay uses base e with this k.

Step-by-step solution

Decay factor
$e^{- k t} = e^{- 0.23 (5)} = e^{- 1.15} = 0.3166$
Exerted demand
$B O D_{5} = B O D_{u} (1 - e^{- k t}) = 320 (1 - 0.3166)$
Evaluate
$B O D_{5} = 320 (0.6834) = 219 m g / L$
Remaining
$320 - 219 = 101 m g / L$

Answer: BOD₅ = 219 mg/L; 101 mg/L remains

Why the other options are there

101 mg/L (remaining reported as BOD₅)
320 mg/L (ultimate reported)

Reference: FE Reference Handbook — Environmental — BOD kinetics

Example 2

Sedimentation basin overflow rate

A rectangular clarifier is 25 m long and 8 m wide, treating 6,000 m³/day. Find the surface overflow rate and, for a 3.0 m depth, the detention time.

Given

L = 25 m, W = 8 m
Q = 6,000 m³/day
Depth = 3.0 m

Find

Overflow rate and detention time

Start with the thinking

Overflow rate uses plan area only — depth does not appear.
Detention time uses the full volume.

Step-by-step solution

Plan area
$A = 25 (8) = 200 m^{2}$
Overflow rate
$v_{o} = Q / A = 6, 000/200 = 30.0 m / d a y$
Volume
$V = 200 (3.0) = 600 m^{3}$
Detention time
$t = V / Q = 600/6, 000 = 0.100 d a y$
Result — v_o = 30 m³/(m²·day), t = 2.4 hours

Answer: v_o = 30 m/day; t = 2.4 h

Why the other options are there

10 m/day (depth divided in)
t = 24 h (day-to-hour conversion missed)

Reference: FE Reference Handbook — Environmental — Sedimentation

Example 3

First-order removal in a CSTR versus a plug-flow reactor — Water Treatment

A reactor of volume 3,407 m³ treats 0.65 m³/s carrying 298 mg/L of a contaminant that decays first-order with k = 0.40 h⁻¹. Compute the hydraulic residence time and the effluent concentration if the tank behaves as a CSTR and as a plug-flow reactor.

Given

V = 3,407 m³
Q = 0.65 m³/s
C₀ = 298 mg/L
k = 0.40 h⁻¹

Find

τ, CSTR effluent and PFR effluent

Start with the thinking

The mass balance for steady state is: in − out − reaction = 0.
For the same volume, plug flow always outperforms a single completely mixed tank for first-order kinetics.

Step-by-step solution

Formula
$τ = V / Q$
Substituting
$τ = 3407/0.65 = 5, 242 s = 1.456 h$
Formula (CSTR)
$C = C_{0} / (1 + k τ)$
Substituting
$C = 298/ (1 + 0.40 \times 1.456) = 188.3 m g / L$
Formula (PFR)
$C = C_{0} e^{(} - k τ)$
Substituting
$C = 298 e^{(} - 0.40 \times 1.456) = 166.5 m g / L$
Comparison — plug flow removes 44.1% versus 36.8% for the CSTR

Answer: τ = 1.46 h; C_CSTR = 188.3 mg/L, C_PFR = 166.5 mg/L

Why the other options are there

124.4 mg/L (linear decay assumed)
188.3 mg/L for both reactors

Reference: FE Reference Handbook — Environmental Engineering → Water Treatment

Example 4

Activated sludge F/M ratio, aeration time and sludge production — Water Treatment

An activated sludge plant treats 24,520 m³/d with an influent BOD of 179 mg/L, effluent BOD 25 mg/L, MLSS 3,057 mg/L, and an aeration basin volume of 5,963 m³. With Y = 0.60 mg VSS/mg BOD, k_d = 0.04 d⁻¹ and an SRT of 5 days, compute the F/M ratio, hydraulic detention time and daily sludge production.

Given

Q = 24,520 m³/d
S₀ = 179 mg/L, S = 25 mg/L
X = 3,057 mg/L, V = 5,963 m³
Y = 0.60, k_d = 0.04 d⁻¹, SRT = 5 d

Find

F/M ratio, detention time τ and sludge wasted per day

Start with the thinking

F/M compares the food entering daily with the microbial mass held in the basin; conventional plants run near 0.2–0.5 d⁻¹.
Endogenous decay reduces net sludge yield as the SRT lengthens.

Step-by-step solution

Formula
$F / M = Q S_{0} / (V X)$
Substituting
$F / M = 24520 (179) / (5963 \times 3057) = 0.241 d^{- 1}$
Formula
$τ = V / Q$
Substituting
$τ = 5963/24520 = 0.2432 d = 5.84 h$
Formula
$P_{x} = Y Q (S_{0} - S) / (1 + k_{d} \cdot S R T)$
Substituting
$P_{x} = 0.60 (24520) (179 - 25) / (1 + 0.04 \times 5) m g / d$
Evaluate
$P_{x} = 1, 888 k g V S S / d a y$

Answer: F/M = 0.24 d⁻¹, τ = 5.8 h, sludge = 1,888 kg/d

Why the other options are there

F/M = 1,436 (basin volume omitted)
P_x = 2,266 kg/d (decay term ignored)

Reference: FE Reference Handbook — Environmental Engineering → Water Treatment

Example 5

Circular clarifier overflow rate, detention time and Stokes settling — Water Treatment

A circular clarifier 24 m in diameter and 4.0 m deep treats 7,703 m³/d. Compute the surface overflow rate, the detention time and the weir loading rate, then check whether a 90 µm particle (SG = 2.65) settles out by Stokes' law.

Given

Q = 7,703 m³/d
D = 24 m, depth = 4.0 m
d_p = 90 µm, ρ_s = 2650 kg/m³
μ = 0.001 Pa·s

Find

SOR, detention time, weir loading and the Stokes settling velocity

Start with the thinking

An ideal settling basin removes every particle whose settling velocity exceeds the overflow rate — depth does not matter for removal, only for detention.
Stokes' law applies to the laminar (small particle) regime.

Step-by-step solution

Surface area
$A = (π /4) (24)^{2} = 452.4 m^{2}$
Formula
$S O R = Q / A$
Substituting — SOR = 7703/452.4 = 17.03 m³/m²·d
Formula — τ = V/Q = A·h/Q
Substituting
$τ = 1, 810/7703 = 5.64 h$
Formula
$w e i r l o a d in g = Q / (π D)$
Substituting — 7703/(π × 24) = 102.2 m³/m·d
Formula
$v_{s} = g (ρ_{s} - ρ) d^{2} / (18 μ)$
Substituting
$v_{s} = 9.81 (1650) (9.0 e - 5)^{2} / (18 \times 0.001) = 629.3 m / d$
Compare
$v_{s} = 629.3 m / d > S O R = 17.03 m / d \to p a r t i c l e i sr e m o v e d$

Answer: SOR = 17.0 m/d, τ = 5.6 h, weir loading = 102.2 m³/m·d, v_s = 629.3 m/d

Why the other options are there

SOR = 25.5 m/d (side area used)
v_s computed with the diameter unsquared

Reference: FE Reference Handbook — Environmental Engineering → Water Treatment

Example 6

Chlorine dose, CT value and daily chemical demand — Water Treatment

A water plant treating 30,111 m³/d applies a chlorine dose of 4.5 mg/L against a demand of 3.4 mg/L, with 92 minutes of contact time. Compute the free residual, the CT value, the daily chlorine mass required, and the concentration remaining after a 4.0-log pathogen reduction of an initial 10⁶ organisms/100 mL.

Given

Q = 30,111 m³/d
Dose = 4.5 mg/L
Demand = 3.4 mg/L
t = 92 min
Target = 4.0 log

Find

Residual, CT, kg/day of chlorine and surviving organisms

Start with the thinking

Residual = dose − demand; the residual, not the dose, drives disinfection credit.
Each log of removal divides the surviving organism count by ten.

Step-by-step solution

Formula
$r es i d u a l = d ose - d e man d$
Substituting
$4.5 - 3.4 = 1.10 m g / L$
Formula
$C T = C \times t$
Substituting — CT = 1.10(92) = 101.2 mg·min/L
Formula
$ma ss = d ose \times Q$
Substituting
$4.5 g / m^{3} \times 30111 m^{3} / d \div 1000 = 135.5 k g / d a y$
Log removal
$N = 1 0^{6} /1 0^{4} .0 = 1.00 e + 2 or g ani s m s /100 m L$

Answer: Residual = 1.10 mg/L, CT = 101.2 mg·min/L, 135.5 kg Cl₂/day, survivors 1.0e+2/100 mL

Why the other options are there

CT = 414.0 (dose used instead of residual)
135,500 kg/day (unit conversion missed)

Reference: FE Reference Handbook — Environmental Engineering → Water Treatment

Example 7

Population projection and design water demand — Water Treatment

A city of 128,435 grows at 1.0% per year. Project the population in 16 years by both geometric and arithmetic growth, then compute the average and peak day water demand at 271 L/capita/day.

Given

P₀ = 128,435
i = 1.0%/yr
n = 16 yr
Per capita use = 271 L/cap/d

Find

Projected population and design flows

Start with the thinking

Geometric growth compounds; arithmetic growth adds a fixed increment and always gives a smaller value over long horizons.
Water systems are sized on peak day (and peak hour) flow, never on the average.

Step-by-step solution

Formula (geometric)
$P = P_{0} (1 + i)^{n}$
Substituting
$P = 128435 (1 + 0.010)^{1} 6 = 150, 600 p eo pl e$
Formula (arithmetic)
$P = P_{0} (1 + in)$
Substituting
$P = 128435 (1 + 0.010 \times 16) = 148, 985 p eo pl e$
Formula
$Q_{a} v g = P \times p er c a p i t a u se$
Substituting
$Q_{a} v g = 150, 600 (271) /1000 = 40, 813 m^{3} / d$
Peak day
$Q_{p} e ak = 2.7 \times 40, 813 = 110, 194 m^{3} / d$

Answer: P = 150,600 (geometric) vs 148,985 (arithmetic); Q_avg = 40,813 m³/d, Q_peak = 110,194 m³/d

Why the other options are there

20,550 (growth increment reported as population)
Q sized on the average day only

Reference: FE Reference Handbook — Environmental Engineering → Water Treatment

Example 8

Chronic daily intake, cancer risk and radioactive decay — Water Treatment

Drinking water contains 0.0240 mg/L of a carcinogen. An adult of 79 kg drinks 1.7 L/day. With a cancer slope factor of 0.40 (mg/kg·d)⁻¹, compute the chronic daily intake and the incremental lifetime cancer risk. Also determine the fraction of a radionuclide with a 22-year half-life remaining after 103 years.

Given

C = 0.0240 mg/L
IR = 1.7 L/d
BW = 79 kg
CSF = 0.40 (mg/kg·d)⁻¹
t½ = 22 yr, t = 103 yr

Find

CDI, lifetime risk and the remaining activity fraction

Start with the thinking

CDI normalises exposure to body weight so a dose-response slope can be applied.
A risk above 10⁻⁶ to 10⁻⁴ is typically the regulatory action range.

Step-by-step solution

Formula
$C D I = (C \times I R) / B W$
Substituting — CDI = (0.0240 × 1.7)/79 = 5.165e-4 mg/kg·d
Formula
$R i s k = C D I \times C S F$
Substituting
$R i s k = 5.165 e - 4 (0.40) = 2.07 e - 4$
Interpretation — above the 10⁻⁶–10⁻⁴ risk range
Formula
$N / N_{0} = (½)^{(} t / t ½)$
Substituting
$(½)^{(} 103/22) = 3.896$

Answer: CDI = 5.16e-4 mg/kg·d, risk = 2.07e-4, 3.90% of the radionuclide remains

Why the other options are there

Risk = 0.00960 (body weight and intake ignored)
-134.1% remaining (linear decay)

Reference: FE Reference Handbook — Environmental Engineering → Water Treatment

Example 9

Completely mixed stream blending — Water Treatment

A stream flowing 24.0 cfs at 27.0 mg/L receives a discharge of 8.0 cfs at 200.0 mg/L. Find the fully mixed concentration.

Given

Q₁ = 24.0 cfs, C₁ = 27.0 mg/L
Q₂ = 8.0 cfs, C₂ = 200.0 mg/L

Find

Mixed concentration C

Start with the thinking

Mass in equals mass out at steady state.
Weight by flow, never a simple average.

Step-by-step solution

Mass balance
$Q_{1} C_{1} + Q_{2} C_{2} = (Q_{1} + Q_{2}) C$
Loads
$24.0 (27.0) = 648.0; 8.0 (200.0) = 1, 600$
Total flow
$24.0 + 8.0 = 32.0 c f s$
Solve
$C = (648.0 + 1, 600) /32.0 = 70.25 m g / L$

Answer: C ≈ 70.3 mg/L

Why the other options are there

113.5 mg/L (unweighted average)
227.0 mg/L (concentrations added)

Reference: FE Reference Handbook — Environmental Engineering → Water Treatment

Example 10

BOD exerted after t days — Water Treatment

A wastewater has ultimate BOD L₀ = 174 mg/L and k = 0.20 /day (base e). How much BOD is exerted in 10 days?

Given

L₀ = 174 mg/L
k = 0.20 /day
t = 10 days

Find

BOD_t

Start with the thinking

BOD exertion is first-order and approaches L₀ asymptotically.
Check whether k is base e or base 10.

Step-by-step solution

First-order
$B O D_{t} = L_{0} (1 - e^{(} - k t))$
Exponent
$k t = 0.20 (10) = 2.00$
Exponential
$e^{(} - 2.00) = 0.1353$
Substituting
$B O D_{t} = 174 (1 - 0.1353) = 150.5 m g / L$
Remaining
$L_{0} - B O D_{t} = 23.5 m g / L$

Answer: BOD_10 ≈ 150.5 mg/L

Why the other options are there

24 mg/L (remaining reported as exerted)
348.0 mg/L (linear decay assumed)

Reference: FE Reference Handbook — Environmental Engineering → Water Treatment

Self-check

Answer these without notes before moving on.

Without looking, state the relation on this page whose left-hand side is the quantity most often requested, and name every symbol in it.
Which assumption, if violated, makes the main relation of this section invalid?
Given a treatment unit or receiving water body, what is the first quantity you would compute, and why that one first?
Which unit conversion in this subject most often produces a wrong answer choice, and what is its numerical factor?
Rework Example 1 above from the givens alone, without reading the solution lines.

Chapter summary

Water Treatment contains 0 relations; you must be able to find this page in under 15 seconds.
Exam style: a mass balance across one reactor or one unit process.
Unit rule: mg/L × MGD × 8.34 = lb/day is the single most used conversion.
Work the 10 examples until the solution path, not the answer, is automatic.

Common traps in this section

mg/L × MGD × 8.34 = lb/day is the single most used conversion
Answering the intermediate quantity instead of the quantity requested.
Rounding intermediate values before the final step.
Using a relation from an adjacent handbook section that shares a symbol.
Skipping the sketch — most lost points on this page start with a misread geometry.