Rapid Mix and Flocculator Design

Environmental Engineering · FE Reference Handbook section

Environmental Engineering

9 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 Rapid Mix and Flocculator Design 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 rapid mix and flocculator design describes physically and when it applies.
State every one of the 9 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. Rapid Mix and Flocculator Design 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: rapid mix and flocculator design.
Capstone Studio instructional photograph

Environmental Engineering — Rapid Mix and Flocculator Design: 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 9 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

Gt	Quantity produced by "Gt = 104 to 105" — read its definition and unit from the handbook line directly above the equation.
G	Quantity produced by "G = root mean square velocity gradient (mixing intensity) [ft/(sec-ft) or m/(s•m)]" — read its definition and unit from the handbook line directly above the equation.
P	Quantity produced by "P = power to the fluid (ft-lb/sec or N•m/s)" — read its definition and unit from the handbook line directly above the equation.
V	Quantity produced by "V = volume (ft3 or m3)" — read its definition and unit from the handbook line directly above the equation.
µ	Quantity produced by "µ = dynamic viscosity [lb/(ft-sec) or Pa•s]" — read its definition and unit from the handbook line directly above the equation.
γ	Quantity produced by "γ = specific weight of water (lb/ft3 or N/m3)" — read its definition and unit from the handbook line directly above the equation.
HL	Quantity produced by "HL = head loss (ft or m)" — read its definition and unit from the handbook line directly above the equation.
t	Quantity produced by "t = time (sec or s)" — 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.

P cHL
where

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

Completely mixed stream blending — Rapid Mix and Flocculator Design

A stream flowing 19.5 cfs at 8.5 mg/L receives a discharge of 6.5 cfs at 295.0 mg/L. Find the fully mixed concentration.

Given

Q₁ = 19.5 cfs, C₁ = 8.5 mg/L
Q₂ = 6.5 cfs, C₂ = 295.0 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
$19.5 (8.5) = 165.8; 6.5 (295.0) = 1, 918$
Total flow
$19.5 + 6.5 = 26.0 c f s$
Solve
$C = (165.8 + 1, 918) /26.0 = 80.13 m g / L$

Answer: C ≈ 80.1 mg/L

Why the other options are there

151.8 mg/L (unweighted average)
303.5 mg/L (concentrations added)

Reference: FE Reference Handbook — Environmental Engineering → Rapid Mix and Flocculator Design

Example 2

Activated sludge F/M ratio, aeration time and sludge production — Rapid Mix and Flocculator Design

An activated sludge plant treats 14,652 m³/d with an influent BOD of 236 mg/L, effluent BOD 23 mg/L, MLSS 3,070 mg/L, and an aeration basin volume of 5,131 m³. With Y = 0.45 mg VSS/mg BOD, k_d = 0.07 d⁻¹ and an SRT of 14 days, compute the F/M ratio, hydraulic detention time and daily sludge production.

Given

Q = 14,652 m³/d
S₀ = 236 mg/L, S = 23 mg/L
X = 3,070 mg/L, V = 5,131 m³
Y = 0.45, k_d = 0.07 d⁻¹, SRT = 14 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
$F / M = 14652 (236) / (5131 \times 3070) = 0.220 d^{- 1}$
Formula
$τ = V / Q$
Substituting
$τ = 5131/14652 = 0.3502 d = 8.40 h$
Formula
$P_{x} = Y Q (S_{0} - S) / (1 + k_{d} \cdot S R T)$
Substituting
$P_{x} = 0.45 (14652) (236 - 23) / (1 + 0.07 \times 14) m g / d$
Evaluate
$P_{x} = 709.3 k g V S S / d a y$

Answer: F/M = 0.22 d⁻¹, τ = 8.4 h, sludge = 709.3 kg/d

Why the other options are there

F/M = 1,126 (basin volume omitted)
P_x = 1,404 kg/d (decay term ignored)

Reference: FE Reference Handbook — Environmental Engineering → Rapid Mix and Flocculator Design

Example 3

Circular clarifier overflow rate, detention time and Stokes settling — Rapid Mix and Flocculator Design

A circular clarifier 31 m in diameter and 5.0 m deep treats 17,744 m³/d. Compute the surface overflow rate, the detention time and the weir loading rate, then check whether a 30 µm particle (SG = 2.65) settles out by Stokes' law.

Given

Q = 17,744 m³/d
D = 31 m, depth = 5.0 m
d_p = 30 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
$A = (π /4) (31)^{2} = 754.8 m^{2}$
Formula
$S O R = Q / A$
Substituting — SOR = 17744/754.8 = 23.51 m³/m²·d
Formula — τ = V/Q = A·h/Q
Substituting
$τ = 3, 774/17744 = 5.10 h$
Formula
$w e i r l o a d in g = Q / (π D)$
Substituting — 17744/(π × 31) = 182.2 m³/m·d
Formula
$v_{s} = g (ρ_{s} - ρ) d^{2} / (18 μ)$
Substituting
$v_{s} = 9.81 (1650) (3.0 e - 5)^{2} / (18 \times 0.001) = 69.93 m / d$
Compare
$v_{s} = 69.93 m / d > S O R = 23.51 m / d \to p a r t i c l e i sr e m o v e d$

Answer: SOR = 23.5 m/d, τ = 5.1 h, weir loading = 182.2 m³/m·d, v_s = 69.9 m/d

Why the other options are there

SOR = 36.4 m/d (side area used)
v_s computed with the diameter unsquared

Reference: FE Reference Handbook — Environmental Engineering → Rapid Mix and Flocculator Design

Example 4

Series particulate control: overall collection efficiency and emission rate — Rapid Mix and Flocculator Design

A stack gas stream of 25 m³/s carries 7 g/m³ of particulate. It passes a cyclone at 85% efficiency followed by a fabric filter at 95.5% efficiency. Compute the concentration after each device, the overall efficiency, and the emission rate in kg/h.

Given

Q = 25 m³/s
C_in = 7 g/m³
η₁ = 0.85
η₂ = 0.955

Find

Intermediate and final concentrations, overall η and kg/h emitted

Start with the thinking

Efficiencies in series multiply as penetrations (1 − η), they never simply add.
The overall penetration is the product of the individual penetrations.

Step-by-step solution

Formula
$C_{1} = C_{i} n (1 - η_{1})$
Substituting
$C_{1} = 7 (1 - 0.85) = 1.050 g / m^{3}$
Formula
$C_{2} = C_{1} (1 - η_{2})$
Substituting
$C_{2} = 1.050 (1 - 0.955) = 0.0473 g / m^{3}$
Formula
$η_{o} v er a l l = 1 - (1 - η_{1}) (1 - η_{2})$
Substituting
$η = 1 - 0.150 \times 0.0450 = 0.99325 = 99.325$
Formula
$e mi ss i o n = C_{2} Q$
Substituting
$0.0473 g / m^{3} \times 25 m^{3} / s \times 3600 s / h \div 1000 = 4.253 k g / h$

Answer: C_out = 0.0473 g/m³, η = 99.33%, emission = 4.25 kg/h

Why the other options are there

η = 180.5% (efficiencies added)
630.0 kg/h (uncontrolled rate)

Reference: FE Reference Handbook — Environmental Engineering → Rapid Mix and Flocculator Design

Example 5

Landfill volume required for a community's solid waste — Rapid Mix and Flocculator Design

A community of 26,938 people generates 2.6 kg/person/day of MSW and diverts 20% through recycling. Compacted in place at 773 kg/m³ with 10% additional daily cover volume, find the airspace needed for 12 years.

Given

Population = 26,938
Generation = 2.6 kg/cap/d
Diversion = 0.20
Compacted density = 773 kg/m³
Cover = 10%, design life 12 yr

Find

Annual and total landfill airspace

Start with the thinking

Only the landfilled fraction consumes airspace — diverted material is subtracted first.
Daily cover soil is real volume and must be added to the waste volume.

Step-by-step solution

Formula
$l an df i l l e d ma ss = p o p u l a t i o n \times r a t e \times (1 - d i v er s i o n)$
Substituting
$26938 (2.6) (1 - 0.20) = 56, 031 k g / d a y$
Formula
$V = ma ss / d e n s i t y$
Substituting
$V = 56, 031/773 = 72.5 m^{3} / d a y$
Annual with cover
$72.5 \times 365 \times 1.10 = 29, 103 m^{3} / y r$
Design life
$29, 103 \times 12 = 349, 234 m^{3}$

Answer: ≈ 29,103 m³/yr, or 349,234 m³ over 12 years

Why the other options are there

396,856 m³ (diversion and cover ignored)
245,415,955 m³ (mass reported as volume)

Reference: FE Reference Handbook — Environmental Engineering → Rapid Mix and Flocculator Design

Example 6

Chlorine dose, CT value and daily chemical demand — Rapid Mix and Flocculator Design

A water plant treating 13,030 m³/d applies a chlorine dose of 5.5 mg/L against a demand of 2.0 mg/L, with 27 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 = 13,030 m³/d
Dose = 5.5 mg/L
Demand = 2.0 mg/L
t = 27 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
$5.5 - 2.0 = 3.50 m g / L$
Formula
$C T = C \times t$
Substituting — CT = 3.50(27) = 94.5 mg·min/L
Formula
$ma ss = d ose \times Q$
Substituting
$5.5 g / m^{3} \times 13030 m^{3} / d \div 1000 = 71.67 k g / d a y$
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 = 3.50 mg/L, CT = 95 mg·min/L, 71.7 kg Cl₂/day, survivors 1.0e+2/100 mL

Why the other options are there

CT = 148.5 (dose used instead of residual)
71,665 kg/day (unit conversion missed)

Reference: FE Reference Handbook — Environmental Engineering → Rapid Mix and Flocculator Design

Example 7

Population projection and design water demand — Rapid Mix and Flocculator Design

A city of 94,060 grows at 2.1% per year. Project the population in 25 years by both geometric and arithmetic growth, then compute the average and peak day water demand at 330 L/capita/day.

Given

P₀ = 94,060
i = 2.1%/yr
n = 25 yr
Per capita use = 330 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
$P = 94060 (1 + 0.021)^{2} 5 = 158, 142 p eo pl e$
Formula (arithmetic)
$P = P_{0} (1 + in)$
Substituting
$P = 94060 (1 + 0.021 \times 25) = 143, 442 p eo pl e$
Formula
$Q_{a} v g = P \times p er c a p i t a u se$
Substituting
$Q_{a} v g = 158, 142 (330) /1000 = 52, 187 m^{3} / d$
Peak day
$Q_{p} e ak = 2.5 \times 52, 187 = 130, 468 m^{3} / d$

Answer: P = 158,142 (geometric) vs 143,442 (arithmetic); Q_avg = 52,187 m³/d, Q_peak = 130,468 m³/d

Why the other options are there

49,382 (growth increment reported as population)
Q sized on the average day only

Reference: FE Reference Handbook — Environmental Engineering → Rapid Mix and Flocculator Design

Example 8

Chronic daily intake, cancer risk and radioactive decay — Rapid Mix and Flocculator Design

Drinking water contains 0.0360 mg/L of a carcinogen. An adult of 74 kg drinks 2.2 L/day. With a cancer slope factor of 1.13 (mg/kg·d)⁻¹, compute the chronic daily intake and the incremental lifetime cancer risk. Also determine the fraction of a radionuclide with a 7-year half-life remaining after 102 years.

Given

C = 0.0360 mg/L
IR = 2.2 L/d
BW = 74 kg
CSF = 1.13 (mg/kg·d)⁻¹
t½ = 7 yr, t = 102 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 = (0.0360 × 2.2)/74 = 1.070e-3 mg/kg·d
Formula
$R i s k = C D I \times C S F$
Substituting
$R i s k = 1.070 e - 3 (1.13) = 1.21 e - 3$
Interpretation — above the 10⁻⁶–10⁻⁴ risk range
Formula
$N / N_{0} = (½)^{(} t / t ½)$
Substituting
$(½)^{(} 102/7) = 0.004$

Answer: CDI = 1.07e-3 mg/kg·d, risk = 1.21e-3, 0.00% of the radionuclide remains

Why the other options are there

Risk = 0.04068 (body weight and intake ignored)
-628.6% remaining (linear decay)

Reference: FE Reference Handbook — Environmental Engineering → Rapid Mix and Flocculator Design

Example 9

BOD exerted after t days — Rapid Mix and Flocculator Design

A wastewater has ultimate BOD L₀ = 181 mg/L and k = 0.28 /day (base e). How much BOD is exerted in 9 days?

Given

L₀ = 181 mg/L
k = 0.28 /day
t = 9 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
$k t = 0.28 (9) = 2.52$
Exponential
$e^{(} - 2.52) = 0.0805$
Substituting
$B O D_{t} = 181 (1 - 0.0805) = 166.4 m g / L$
Remaining
$L_{0} - B O D_{t} = 14.6 m g / L$

Answer: BOD_9 ≈ 166.4 mg/L

Why the other options are there

15 mg/L (remaining reported as exerted)
456.1 mg/L (linear decay assumed)

Reference: FE Reference Handbook — Environmental Engineering → Rapid Mix and Flocculator Design

Example 10

Hydraulic detention time in a tank — Rapid Mix and Flocculator Design

A treatment tank holds 64,541 gal and treats 2.4 MGD. Find the hydraulic detention time.

Given

V = 64,541 gal
Q = 2.4 MGD

Find

Detention time θ

Start with the thinking

θ = V/Q with consistent volume units.
Express the answer in hours for design comparison.

Step-by-step solution

Detention
$θ = V / Q$
Flow in gal/day
$2.4 \times 1 0^{6} = 2, 400, 000 g a l / d a y$
Substituting
$θ = 64541/2, 400, 000 = 0.0269 d a y$
Convert
$θ = 0.0269 (24) = 0.65 h r$

Answer: θ ≈ 0.65 hr

Why the other options are there

0.027 hr (day/hour conversion missed)
37.19 hr (ratio inverted)

Reference: FE Reference Handbook — Environmental Engineering → Rapid Mix and Flocculator Design

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

Rapid Mix and Flocculator Design contains 9 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.