Daughter Product Maximum Activity Time

Environmental Engineering · FE Reference Handbook section

Environmental Engineering

1 formulas

10 exam-style examples

~47 min

All Environmental Engineering lectures

Learning objectives

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

This chapter section covers Daughter Product Maximum Activity Time 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 daughter product maximum activity time describes physically and when it applies.
State every one of the 1 relation 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. Daughter Product Maximum Activity Time 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: daughter product maximum activity time.
Capstone Studio instructional photograph

Environmental Engineering — Daughter Product Maximum Activity Time: 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 1 relation 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.

ln m2 − ln m1
m2 − m1

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

Activated sludge F/M ratio, aeration time and sludge production — Daughter Product Maximum Activity Time

An activated sludge plant treats 15,146 m³/d with an influent BOD of 309 mg/L, effluent BOD 5 mg/L, MLSS 3,131 mg/L, and an aeration basin volume of 5,411 m³. With Y = 0.65 mg VSS/mg BOD, k_d = 0.07 d⁻¹ and an SRT of 8 days, compute the F/M ratio, hydraulic detention time and daily sludge production.

Given

Q = 15,146 m³/d
S₀ = 309 mg/L, S = 5 mg/L
X = 3,131 mg/L, V = 5,411 m³
Y = 0.65, k_d = 0.07 d⁻¹, SRT = 8 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
$F / M = 15146 (309) / (5411 \times 3131) = 0.276 d^{- 1}$
Formula
$τ = V / Q$
Substituting
$τ = 5411/15146 = 0.3573 d = 8.57 h$
Formula
$P_{x} = Y Q (S_{0} - S) / (1 + k_{d} \cdot S R T)$
Substituting
$P_{x} = 0.65 (15146) (309 - 5) / (1 + 0.07 \times 8) m g / d$
Evaluate
$P_{x} = 1, 918 k g V S S / d a y$

Answer: F/M = 0.28 d⁻¹, τ = 8.6 h, sludge = 1,918 kg/d

Why the other options are there

F/M = 1,495 (basin volume omitted)
P_x = 2,993 kg/d (decay term ignored)

Reference: FE Reference Handbook — Environmental Engineering → Daughter Product Maximum Activity Time

Example 2

Chronic daily intake, cancer risk and radioactive decay — Daughter Product Maximum Activity Time

Drinking water contains 0.0060 mg/L of a carcinogen. An adult of 69 kg drinks 2.2 L/day. With a cancer slope factor of 1.63 (mg/kg·d)⁻¹, compute the chronic daily intake and the incremental lifetime cancer risk. Also determine the fraction of a radionuclide with a 29-year half-life remaining after 100 years.

Given

C = 0.0060 mg/L
IR = 2.2 L/d
BW = 69 kg
CSF = 1.63 (mg/kg·d)⁻¹
t½ = 29 yr, t = 100 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 = (0.0060 × 2.2)/69 = 1.913e-4 mg/kg·d
Formula
$R i s k = C D I \times C S F$
Substituting
$R i s k = 1.913 e - 4 (1.63) = 3.12 e - 4$
Interpretation — above the 10⁻⁶–10⁻⁴ risk range
Formula
$N / N_{0} = (½)^{(} t / t ½)$
Substituting
$(½)^{(} 100/29) = 9.161$

Answer: CDI = 1.91e-4 mg/kg·d, risk = 3.12e-4, 9.16% of the radionuclide remains

Why the other options are there

Risk = 0.00978 (body weight and intake ignored)
-72.4% remaining (linear decay)

Reference: FE Reference Handbook — Environmental Engineering → Daughter Product Maximum Activity Time

Example 3

Activated sludge F/M ratio, aeration time and sludge production — Daughter Product Maximum Activity Time (2)

An activated sludge plant treats 21,568 m³/d with an influent BOD of 265 mg/L, effluent BOD 9 mg/L, MLSS 2,260 mg/L, and an aeration basin volume of 1,261 m³. With Y = 0.70 mg VSS/mg BOD, k_d = 0.06 d⁻¹ and an SRT of 15 days, compute the F/M ratio, hydraulic detention time and daily sludge production.

Given

Q = 21,568 m³/d
S₀ = 265 mg/L, S = 9 mg/L
X = 2,260 mg/L, V = 1,261 m³
Y = 0.70, k_d = 0.06 d⁻¹, SRT = 15 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
$F / M = 21568 (265) / (1261 \times 2260) = 2.006 d^{- 1}$
Formula
$τ = V / Q$
Substituting
$τ = 1261/21568 = 0.0585 d = 1.40 h$
Formula
$P_{x} = Y Q (S_{0} - S) / (1 + k_{d} \cdot S R T)$
Substituting
$P_{x} = 0.70 (21568) (265 - 9) / (1 + 0.06 \times 15) m g / d$
Evaluate
$P_{x} = 2, 034 k g V S S / d a y$

Answer: F/M = 2.01 d⁻¹, τ = 1.4 h, sludge = 2,034 kg/d

Why the other options are there

F/M = 2,529 (basin volume omitted)
P_x = 3,865 kg/d (decay term ignored)

Reference: FE Reference Handbook — Environmental Engineering → Daughter Product Maximum Activity Time

Example 4

Chronic daily intake, cancer risk and radioactive decay — Daughter Product Maximum Activity Time (2)

Drinking water contains 0.0080 mg/L of a carcinogen. An adult of 84 kg drinks 2.2 L/day. With a cancer slope factor of 1.48 (mg/kg·d)⁻¹, compute the chronic daily intake and the incremental lifetime cancer risk. Also determine the fraction of a radionuclide with a 5-year half-life remaining after 117 years.

Given

C = 0.0080 mg/L
IR = 2.2 L/d
BW = 84 kg
CSF = 1.48 (mg/kg·d)⁻¹
t½ = 5 yr, t = 117 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 = (0.0080 × 2.2)/84 = 2.095e-4 mg/kg·d
Formula
$R i s k = C D I \times C S F$
Substituting
$R i s k = 2.095 e - 4 (1.48) = 3.10 e - 4$
Interpretation — above the 10⁻⁶–10⁻⁴ risk range
Formula
$N / N_{0} = (½)^{(} t / t ½)$
Substituting
$(½)^{(} 117/5) = 0.000$

Answer: CDI = 2.10e-4 mg/kg·d, risk = 3.10e-4, 0.00% of the radionuclide remains

Why the other options are there

Risk = 0.01184 (body weight and intake ignored)
-1,070% remaining (linear decay)

Reference: FE Reference Handbook — Environmental Engineering → Daughter Product Maximum Activity Time

Example 5

Activated sludge F/M ratio, aeration time and sludge production — Daughter Product Maximum Activity Time (3)

An activated sludge plant treats 18,496 m³/d with an influent BOD of 369 mg/L, effluent BOD 7 mg/L, MLSS 2,391 mg/L, and an aeration basin volume of 3,970 m³. With Y = 0.55 mg VSS/mg BOD, k_d = 0.07 d⁻¹ and an SRT of 17 days, compute the F/M ratio, hydraulic detention time and daily sludge production.

Given

Q = 18,496 m³/d
S₀ = 369 mg/L, S = 7 mg/L
X = 2,391 mg/L, V = 3,970 m³
Y = 0.55, k_d = 0.07 d⁻¹, SRT = 17 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
$F / M = 18496 (369) / (3970 \times 2391) = 0.719 d^{- 1}$
Formula
$τ = V / Q$
Substituting
$τ = 3970/18496 = 0.2146 d = 5.15 h$
Formula
$P_{x} = Y Q (S_{0} - S) / (1 + k_{d} \cdot S R T)$
Substituting
$P_{x} = 0.55 (18496) (369 - 7) / (1 + 0.07 \times 17) m g / d$
Evaluate
$P_{x} = 1, 682 k g V S S / d a y$

Answer: F/M = 0.72 d⁻¹, τ = 5.2 h, sludge = 1,682 kg/d

Why the other options are there

F/M = 2,854 (basin volume omitted)
P_x = 3,683 kg/d (decay term ignored)

Reference: FE Reference Handbook — Environmental Engineering → Daughter Product Maximum Activity Time

Example 6

Chronic daily intake, cancer risk and radioactive decay — Daughter Product Maximum Activity Time (3)

Drinking water contains 0.0360 mg/L of a carcinogen. An adult of 64 kg drinks 1.5 L/day. With a cancer slope factor of 0.45 (mg/kg·d)⁻¹, compute the chronic daily intake and the incremental lifetime cancer risk. Also determine the fraction of a radionuclide with a 49-year half-life remaining after 30 years.

Given

C = 0.0360 mg/L
IR = 1.5 L/d
BW = 64 kg
CSF = 0.45 (mg/kg·d)⁻¹
t½ = 49 yr, t = 30 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 × 1.5)/64 = 8.437e-4 mg/kg·d
Formula
$R i s k = C D I \times C S F$
Substituting
$R i s k = 8.437 e - 4 (0.45) = 3.80 e - 4$
Interpretation — above the 10⁻⁶–10⁻⁴ risk range
Formula
$N / N_{0} = (½)^{(} t / t ½)$
Substituting
$(½)^{(} 30/49) = 65.418$

Answer: CDI = 8.44e-4 mg/kg·d, risk = 3.80e-4, 65.42% of the radionuclide remains

Why the other options are there

Risk = 0.01620 (body weight and intake ignored)
69.4% remaining (linear decay)

Reference: FE Reference Handbook — Environmental Engineering → Daughter Product Maximum Activity Time

Example 7

Activated sludge F/M ratio, aeration time and sludge production — Daughter Product Maximum Activity Time (4)

An activated sludge plant treats 13,583 m³/d with an influent BOD of 218 mg/L, effluent BOD 16 mg/L, MLSS 3,753 mg/L, and an aeration basin volume of 5,120 m³. With Y = 0.40 mg VSS/mg BOD, k_d = 0.05 d⁻¹ and an SRT of 16 days, compute the F/M ratio, hydraulic detention time and daily sludge production.

Given

Q = 13,583 m³/d
S₀ = 218 mg/L, S = 16 mg/L
X = 3,753 mg/L, V = 5,120 m³
Y = 0.40, k_d = 0.05 d⁻¹, SRT = 16 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
$F / M = 13583 (218) / (5120 \times 3753) = 0.154 d^{- 1}$
Formula
$τ = V / Q$
Substituting
$τ = 5120/13583 = 0.3769 d = 9.05 h$
Formula
$P_{x} = Y Q (S_{0} - S) / (1 + k_{d} \cdot S R T)$
Substituting
$P_{x} = 0.40 (13583) (218 - 16) / (1 + 0.05 \times 16) m g / d$
Evaluate
$P_{x} = 609.7 k g V S S / d a y$

Answer: F/M = 0.15 d⁻¹, τ = 9.0 h, sludge = 609.7 kg/d

Why the other options are there

F/M = 789.0 (basin volume omitted)
P_x = 1,098 kg/d (decay term ignored)

Reference: FE Reference Handbook — Environmental Engineering → Daughter Product Maximum Activity Time

Example 8

Chronic daily intake, cancer risk and radioactive decay — Daughter Product Maximum Activity Time (4)

Drinking water contains 0.0540 mg/L of a carcinogen. An adult of 80 kg drinks 2.0 L/day. With a cancer slope factor of 0.62 (mg/kg·d)⁻¹, compute the chronic daily intake and the incremental lifetime cancer risk. Also determine the fraction of a radionuclide with a 50-year half-life remaining after 78 years.

Given

C = 0.0540 mg/L
IR = 2.0 L/d
BW = 80 kg
CSF = 0.62 (mg/kg·d)⁻¹
t½ = 50 yr, t = 78 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 = (0.0540 × 2.0)/80 = 1.350e-3 mg/kg·d
Formula
$R i s k = C D I \times C S F$
Substituting
$R i s k = 1.350 e - 3 (0.62) = 8.37 e - 4$
Interpretation — above the 10⁻⁶–10⁻⁴ risk range
Formula
$N / N_{0} = (½)^{(} t / t ½)$
Substituting
$(½)^{(} 78/50) = 33.915$

Answer: CDI = 1.35e-3 mg/kg·d, risk = 8.37e-4, 33.92% of the radionuclide remains

Why the other options are there

Risk = 0.03348 (body weight and intake ignored)
22.0% remaining (linear decay)

Reference: FE Reference Handbook — Environmental Engineering → Daughter Product Maximum Activity Time

Example 9

Circular clarifier overflow rate, detention time and Stokes settling — Daughter Product Maximum Activity Time

A circular clarifier 27 m in diameter and 4.0 m deep treats 8,218 m³/d. Compute the surface overflow rate, the detention time and the weir loading rate, then check whether a 70 µm particle (SG = 2.65) settles out by Stokes' law.

Given

Q = 8,218 m³/d
D = 27 m, depth = 4.0 m
d_p = 70 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
$A = (π /4) (27)^{2} = 572.6 m^{2}$
Formula
$S O R = Q / A$
Substituting — SOR = 8218/572.6 = 14.35 m³/m²·d
Formula — τ = V/Q = A·h/Q
Substituting
$τ = 2, 290/8218 = 6.69 h$
Formula
$w e i r l o a d in g = Q / (π D)$
Substituting — 8218/(π × 27) = 96.9 m³/m·d
Formula
$v_{s} = g (ρ_{s} - ρ) d^{2} / (18 μ)$
Substituting
$v_{s} = 9.81 (1650) (7.0 e - 5)^{2} / (18 \times 0.001) = 380.7 m / d$
Compare
$v_{s} = 380.7 m / d > S O R = 14.35 m / d \to p a r t i c l e i sr e m o v e d$

Answer: SOR = 14.4 m/d, τ = 6.7 h, weir loading = 97 m³/m·d, v_s = 380.7 m/d

Why the other options are there

SOR = 24.2 m/d (side area used)
v_s computed with the diameter unsquared

Reference: FE Reference Handbook — Environmental Engineering → Daughter Product Maximum Activity Time

Example 10

Series particulate control: overall collection efficiency and emission rate — Daughter Product Maximum Activity Time

A stack gas stream of 51 m³/s carries 23 g/m³ of particulate. It passes a cyclone at 82% efficiency followed by a fabric filter at 94.5% efficiency. Compute the concentration after each device, the overall efficiency, and the emission rate in kg/h.

Given

Q = 51 m³/s
C_in = 23 g/m³
η₁ = 0.82
η₂ = 0.945

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
$C_{1} = 23 (1 - 0.82) = 4.140 g / m^{3}$
Formula
$C_{2} = C_{1} (1 - η_{2})$
Substituting
$C_{2} = 4.140 (1 - 0.945) = 0.2277 g / m^{3}$
Formula
$η_{o} v er a l l = 1 - (1 - η_{1}) (1 - η_{2})$
Substituting
$η = 1 - 0.180 \times 0.0550 = 0.99010 = 99.010$
Formula
$e mi ss i o n = C_{2} Q$
Substituting
$0.2277 g / m^{3} \times 51 m^{3} / s \times 3600 s / h \div 1000 = 41.806 k g / h$

Answer: C_out = 0.2277 g/m³, η = 99.01%, emission = 41.81 kg/h

Why the other options are there

η = 176.5% (efficiencies added)
4,223 kg/h (uncontrolled rate)

Reference: FE Reference Handbook — Environmental Engineering → Daughter Product Maximum Activity Time

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

Daughter Product Maximum Activity Time contains 1 relation; 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.