Disinfection

Environmental Engineering · FE Reference Handbook section

Environmental Engineering

8 formulas

10 exam-style examples

~60 min

All Environmental Engineering lectures

Learning objectives

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

This chapter section covers Disinfection 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 disinfection describes physically and when it applies.
State every one of the 8 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. Disinfection 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: disinfection.
Capstone Studio instructional photograph

Environmental Engineering — Disinfection: 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 8 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.

Notation used in this section

CTcalc	Quantity produced by "CTcalc = C × t10" — read its definition and unit from the handbook line directly above the equation.
C	Quantity produced by "C = residual disinfectant concentration measured during peak hourly flow (mg/L)" — read its definition and unit from the handbook line directly above the equation.
t10	Quantity produced by "t10 = time it takes 10% of the water to flow through the reactor measured during peak hourly flow (min)" — read its definition and unit from the handbook line directly above the equation.
θ	Quantity produced by "θ = hydraulic residence time (min)" — read its definition and unit from the handbook line directly above the equation.
BF	Quantity produced by "BF = baffling factor" — read its definition and unit from the handbook line directly above the equation.

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

where
Adapted from Guidance Manual LT1ESWTR Disinfection Profiling and Benchmarking, U.S. Environmental Protection Agency, 2003.
Baffling Factors
Baffling Baffling
Baffling Description
Condition Factor
Unbaffled 0.1 None, agitated basin, very low
(mixed flow) length to width ratio, high inlet
and outlet flow velocities.
Poor 0.3 Single or multiple unbaffled
inlets and outlets, no intra-basin
baffles.
Average 0.5 Baffled inlet or outlet with some
intra-basin baffles.
Superior 0.7 Perforated inlet baffle,
serpentine or perforated
intra-basin baffles, outlet
weir or perforated launders.
Perfect 1.0 Very high length to width ratio
(plug flow) (pipeline flow), perforated inlet,
outlet, and intra-basin baffles.
Guidance Manual LT1ESWTR Disinfection Profiling and Benchmarking, U.S. Environmental Protection Agency, 2003.

Core formulas for this FE topic

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

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

Chemical feed rate from dosage — Disinfection

A plant treating 5.8 MGD requires a 3.5 mg/L dose. What is the daily chemical demand in pounds?

Given

Q = 5.8 MGD
Dose = 3.5 mg/L
8.34 lb/gal

Find

lb/day of chemical

Start with the thinking

The 8.34 factor converts mg/L·MGD to lb/day.
Adjust for purity if the product is not 100% active.

Step-by-step solution

Feed
$l b / d a y = Q (M G D) \times d ose (m g / L) \times 8.34$
Substituting
$l b / d a y = 5.8 (3.5) (8.34)$
Evaluate — 169.3 lb/day
At 65% available strength — 260.5 lb/day of product

Answer: ≈ 169.3 lb/day

Why the other options are there

20.3 lb/day (8.34 omitted)
2.43 lb/day (factor divided)

Reference: FE Reference Handbook — Environmental Engineering → Disinfection

Example 4

Chlorine dose, CT value and daily chemical demand — Disinfection

A water plant treating 19,812 m³/d applies a chlorine dose of 2.0 mg/L against a demand of 1.4 mg/L, with 95 minutes of contact time. Compute the free residual, the CT value, the daily chlorine mass required, and the concentration remaining after a 2.0-log pathogen reduction of an initial 10⁶ organisms/100 mL.

Given

Q = 19,812 m³/d
Dose = 2.0 mg/L
Demand = 1.4 mg/L
t = 95 min
Target = 2.0 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
$2.0 - 1.4 = 0.60 m g / L$
Formula
$C T = C \times t$
Substituting — CT = 0.60(95) = 57.0 mg·min/L
Formula
$ma ss = d ose \times Q$
Substituting
$2.0 g / m^{3} \times 19812 m^{3} / d \div 1000 = 39.62 k g / d a y$
Log removal
$N = 1 0^{6} /1 0^{2} .0 = 1.00 e + 4 or g ani s m s /100 m L$

Answer: Residual = 0.60 mg/L, CT = 57 mg·min/L, 39.6 kg Cl₂/day, survivors 1.0e+4/100 mL

Why the other options are there

CT = 190.0 (dose used instead of residual)
39,624 kg/day (unit conversion missed)

Reference: FE Reference Handbook — Environmental Engineering → Disinfection

Example 5

Chemical feed rate from dosage — Disinfection (2)

A plant treating 8.0 MGD requires a 11.0 mg/L dose. What is the daily chemical demand in pounds?

Given

Q = 8.0 MGD
Dose = 11.0 mg/L
8.34 lb/gal

Find

lb/day of chemical

Start with the thinking

The 8.34 factor converts mg/L·MGD to lb/day.
Adjust for purity if the product is not 100% active.

Step-by-step solution

Feed
$l b / d a y = Q (M G D) \times d ose (m g / L) \times 8.34$
Substituting
$l b / d a y = 8.0 (11.0) (8.34)$
Evaluate — 733.9 lb/day
At 65% available strength — 1,129 lb/day of product

Answer: ≈ 733.9 lb/day

Why the other options are there

88.0 lb/day (8.34 omitted)
10.55 lb/day (factor divided)

Reference: FE Reference Handbook — Environmental Engineering → Disinfection

Example 6

Chlorine dose, CT value and daily chemical demand — Disinfection (2)

A water plant treating 17,524 m³/d applies a chlorine dose of 2.5 mg/L against a demand of 1.0 mg/L, with 66 minutes of contact time. Compute the free residual, the CT value, the daily chlorine mass required, and the concentration remaining after a 3.5-log pathogen reduction of an initial 10⁶ organisms/100 mL.

Given

Q = 17,524 m³/d
Dose = 2.5 mg/L
Demand = 1.0 mg/L
t = 66 min
Target = 3.5 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
$2.5 - 1.0 = 1.50 m g / L$
Formula
$C T = C \times t$
Substituting — CT = 1.50(66) = 99.0 mg·min/L
Formula
$ma ss = d ose \times Q$
Substituting
$2.5 g / m^{3} \times 17524 m^{3} / d \div 1000 = 43.81 k g / d a y$
Log removal
$N = 1 0^{6} /1 0^{3} .5 = 3.16 e + 2 or g ani s m s /100 m L$

Answer: Residual = 1.50 mg/L, CT = 99 mg·min/L, 43.8 kg Cl₂/day, survivors 3.2e+2/100 mL

Why the other options are there

CT = 165.0 (dose used instead of residual)
43,810 kg/day (unit conversion missed)

Reference: FE Reference Handbook — Environmental Engineering → Disinfection

Example 7

Chemical feed rate from dosage — Disinfection (3)

A plant treating 2.3 MGD requires a 1.5 mg/L dose. What is the daily chemical demand in pounds?

Given

Q = 2.3 MGD
Dose = 1.5 mg/L
8.34 lb/gal

Find

lb/day of chemical

Start with the thinking

The 8.34 factor converts mg/L·MGD to lb/day.
Adjust for purity if the product is not 100% active.

Step-by-step solution

Feed
$l b / d a y = Q (M G D) \times d ose (m g / L) \times 8.34$
Substituting
$l b / d a y = 2.3 (1.5) (8.34)$
Evaluate — 28.8 lb/day
At 65% available strength — 44.3 lb/day of product

Answer: ≈ 29 lb/day

Why the other options are there

3.4 lb/day (8.34 omitted)
0.41 lb/day (factor divided)

Reference: FE Reference Handbook — Environmental Engineering → Disinfection

Example 8

Chlorine dose, CT value and daily chemical demand — Disinfection (3)

A water plant treating 31,866 m³/d applies a chlorine dose of 2.0 mg/L against a demand of 2.6 mg/L, with 101 minutes of contact time. Compute the free residual, the CT value, the daily chlorine mass required, and the concentration remaining after a 2.0-log pathogen reduction of an initial 10⁶ organisms/100 mL.

Given

Q = 31,866 m³/d
Dose = 2.0 mg/L
Demand = 2.6 mg/L
t = 101 min
Target = 2.0 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
$2.0 - 2.6 = 0.20 m g / L$
Formula
$C T = C \times t$
Substituting — CT = 0.20(101) = 20.2 mg·min/L
Formula
$ma ss = d ose \times Q$
Substituting
$2.0 g / m^{3} \times 31866 m^{3} / d \div 1000 = 63.73 k g / d a y$
Log removal
$N = 1 0^{6} /1 0^{2} .0 = 1.00 e + 4 or g ani s m s /100 m L$

Answer: Residual = 0.20 mg/L, CT = 20 mg·min/L, 63.7 kg Cl₂/day, survivors 1.0e+4/100 mL

Why the other options are there

CT = 202.0 (dose used instead of residual)
63,732 kg/day (unit conversion missed)

Reference: FE Reference Handbook — Environmental Engineering → Disinfection

Example 9

Chemical feed rate from dosage — Disinfection (4)

A plant treating 6.7 MGD requires a 7.5 mg/L dose. What is the daily chemical demand in pounds?

Given

Q = 6.7 MGD
Dose = 7.5 mg/L
8.34 lb/gal

Find

lb/day of chemical

Start with the thinking

The 8.34 factor converts mg/L·MGD to lb/day.
Adjust for purity if the product is not 100% active.

Step-by-step solution

Feed
$l b / d a y = Q (M G D) \times d ose (m g / L) \times 8.34$
Substituting
$l b / d a y = 6.7 (7.5) (8.34)$
Evaluate — 419.1 lb/day
At 65% available strength — 644.7 lb/day of product

Answer: ≈ 419.1 lb/day

Why the other options are there

50.3 lb/day (8.34 omitted)
6.03 lb/day (factor divided)

Reference: FE Reference Handbook — Environmental Engineering → Disinfection

Example 10

Chlorine dose, CT value and daily chemical demand — Disinfection (4)

A water plant treating 17,787 m³/d applies a chlorine dose of 2.0 mg/L against a demand of 1.4 mg/L, with 82 minutes of contact time. Compute the free residual, the CT value, the daily chlorine mass required, and the concentration remaining after a 3.0-log pathogen reduction of an initial 10⁶ organisms/100 mL.

Given

Q = 17,787 m³/d
Dose = 2.0 mg/L
Demand = 1.4 mg/L
t = 82 min
Target = 3.0 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
$2.0 - 1.4 = 0.60 m g / L$
Formula
$C T = C \times t$
Substituting — CT = 0.60(82) = 49.2 mg·min/L
Formula
$ma ss = d ose \times Q$
Substituting
$2.0 g / m^{3} \times 17787 m^{3} / d \div 1000 = 35.57 k g / d a y$
Log removal
$N = 1 0^{6} /1 0^{3} .0 = 1.00 e + 3 or g ani s m s /100 m L$

Answer: Residual = 0.60 mg/L, CT = 49 mg·min/L, 35.6 kg Cl₂/day, survivors 1.0e+3/100 mL

Why the other options are there

CT = 164.0 (dose used instead of residual)
35,574 kg/day (unit conversion missed)

Reference: FE Reference Handbook — Environmental Engineering → Disinfection

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

Disinfection contains 8 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.

Disinfection

Learning objectives

Lecture

Theory, developed

Notation used in this section

Handbook notes for this section

Core formulas for this FE topic

Chlorinecontactchamberlength−to−widthratio=20:1to50:1

CTcalc=C×t10

CTcalc=calculatedCTvalue(mg•mm/L)

C=residualdisinfectantconcentrationmeasuredduringpeakhourlyflow(mg/L)

t10=timeittakes10

=canbedeterminedfromtracesstudydataorthefollowingrelationship,t10(approx)=θ×BF

θ=hydraulicresidencetime(min)

BF=bafflingfactor

Worked exam-style examples

Self-check

Chapter summary

Common traps in this section