Methanol Requirement for Biologically Treated Wastewater

Environmental Engineering · FE Reference Handbook section

Environmental Engineering

5 formulas

10 exam-style examples

~55 min

All Environmental Engineering lectures

Learning objectives

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

This chapter section covers Methanol Requirement for Biologically Treated Wastewater 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 methanol requirement for biologically treated wastewater describes physically and when it applies.
State every one of the 5 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. Methanol Requirement for Biologically Treated Wastewater 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: methanol requirement for biologically treated wastewater.
Capstone Studio instructional photograph

Environmental Engineering — Methanol Requirement for Biologically Treated Wastewater: 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 5 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

Cm	Quantity produced by "Cm = 2.47No + 1.53Nl + 0.87Do" — read its definition and unit from the handbook line directly above the equation.
No	Quantity produced by "No = initial nitrate-nitrogen concentration (mg/L)" — read its definition and unit from the handbook line directly above the equation.
Nl	Quantity produced by "Nl = initial nitrite-nitrogen concentration (mg/L)" — read its definition and unit from the handbook line directly above the equation.
Do	Quantity produced by "Do = initial dissolved-oxygen concentration (mg/L)" — 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

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 — Methanol Requirement for Biologically Treated Wastewater

An activated sludge plant treats 2,672 m³/d with an influent BOD of 218 mg/L, effluent BOD 18 mg/L, MLSS 3,218 mg/L, and an aeration basin volume of 945 m³. With Y = 0.55 mg VSS/mg BOD, k_d = 0.05 d⁻¹ and an SRT of 15 days, compute the F/M ratio, hydraulic detention time and daily sludge production.

Given

Q = 2,672 m³/d
S₀ = 218 mg/L, S = 18 mg/L
X = 3,218 mg/L, V = 945 m³
Y = 0.55, k_d = 0.05 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
$F / M = 2672 (218) / (945 \times 3218) = 0.192 d^{- 1}$
Formula
$τ = V / Q$
Substituting
$τ = 945/2672 = 0.3537 d = 8.49 h$
Formula
$P_{x} = Y Q (S_{0} - S) / (1 + k_{d} \cdot S R T)$
Substituting
$P_{x} = 0.55 (2672) (218 - 18) / (1 + 0.05 \times 15) m g / d$
Evaluate
$P_{x} = 168.0 k g V S S / d a y$

Answer: F/M = 0.19 d⁻¹, τ = 8.5 h, sludge = 168.0 kg/d

Why the other options are there

F/M = 181.0 (basin volume omitted)
P_x = 293.9 kg/d (decay term ignored)

Reference: FE Reference Handbook — Environmental Engineering → Methanol Requirement for Biologically Treated Wastewater

Example 2

Activated sludge F/M ratio, aeration time and sludge production — Methanol Requirement for Biologically Treated Wastewater (2)

An activated sludge plant treats 13,714 m³/d with an influent BOD of 329 mg/L, effluent BOD 13 mg/L, MLSS 2,790 mg/L, and an aeration basin volume of 3,836 m³. With Y = 0.65 mg VSS/mg BOD, k_d = 0.05 d⁻¹ and an SRT of 12 days, compute the F/M ratio, hydraulic detention time and daily sludge production.

Given

Q = 13,714 m³/d
S₀ = 329 mg/L, S = 13 mg/L
X = 2,790 mg/L, V = 3,836 m³
Y = 0.65, k_d = 0.05 d⁻¹, SRT = 12 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
$F / M = 13714 (329) / (3836 \times 2790) = 0.422 d^{- 1}$
Formula
$τ = V / Q$
Substituting
$τ = 3836/13714 = 0.2797 d = 6.71 h$
Formula
$P_{x} = Y Q (S_{0} - S) / (1 + k_{d} \cdot S R T)$
Substituting
$P_{x} = 0.65 (13714) (329 - 13) / (1 + 0.05 \times 12) m g / d$
Evaluate
$P_{x} = 1, 761 k g V S S / d a y$

Answer: F/M = 0.42 d⁻¹, τ = 6.7 h, sludge = 1,761 kg/d

Why the other options are there

F/M = 1,617 (basin volume omitted)
P_x = 2,817 kg/d (decay term ignored)

Reference: FE Reference Handbook — Environmental Engineering → Methanol Requirement for Biologically Treated Wastewater

Example 3

Activated sludge F/M ratio, aeration time and sludge production — Methanol Requirement for Biologically Treated Wastewater (3)

An activated sludge plant treats 10,120 m³/d with an influent BOD of 263 mg/L, effluent BOD 10 mg/L, MLSS 1,961 mg/L, and an aeration basin volume of 1,119 m³. With Y = 0.65 mg VSS/mg BOD, k_d = 0.07 d⁻¹ and an SRT of 9 days, compute the F/M ratio, hydraulic detention time and daily sludge production.

Given

Q = 10,120 m³/d
S₀ = 263 mg/L, S = 10 mg/L
X = 1,961 mg/L, V = 1,119 m³
Y = 0.65, k_d = 0.07 d⁻¹, SRT = 9 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
$F / M = 10120 (263) / (1119 \times 1961) = 1.213 d^{- 1}$
Formula
$τ = V / Q$
Substituting
$τ = 1119/10120 = 0.1106 d = 2.65 h$
Formula
$P_{x} = Y Q (S_{0} - S) / (1 + k_{d} \cdot S R T)$
Substituting
$P_{x} = 0.65 (10120) (263 - 10) / (1 + 0.07 \times 9) m g / d$
Evaluate
$P_{x} = 1, 021 k g V S S / d a y$

Answer: F/M = 1.21 d⁻¹, τ = 2.7 h, sludge = 1,021 kg/d

Why the other options are there

F/M = 1,357 (basin volume omitted)
P_x = 1,664 kg/d (decay term ignored)

Reference: FE Reference Handbook — Environmental Engineering → Methanol Requirement for Biologically Treated Wastewater

Example 4

Activated sludge F/M ratio, aeration time and sludge production — Methanol Requirement for Biologically Treated Wastewater (4)

An activated sludge plant treats 15,038 m³/d with an influent BOD of 166 mg/L, effluent BOD 15 mg/L, MLSS 2,183 mg/L, and an aeration basin volume of 4,217 m³. With Y = 0.50 mg VSS/mg BOD, k_d = 0.06 d⁻¹ and an SRT of 9 days, compute the F/M ratio, hydraulic detention time and daily sludge production.

Given

Q = 15,038 m³/d
S₀ = 166 mg/L, S = 15 mg/L
X = 2,183 mg/L, V = 4,217 m³
Y = 0.50, k_d = 0.06 d⁻¹, SRT = 9 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
$F / M = 15038 (166) / (4217 \times 2183) = 0.271 d^{- 1}$
Formula
$τ = V / Q$
Substituting
$τ = 4217/15038 = 0.2804 d = 6.73 h$
Formula
$P_{x} = Y Q (S_{0} - S) / (1 + k_{d} \cdot S R T)$
Substituting
$P_{x} = 0.50 (15038) (166 - 15) / (1 + 0.06 \times 9) m g / d$
Evaluate
$P_{x} = 737.3 k g V S S / d a y$

Answer: F/M = 0.27 d⁻¹, τ = 6.7 h, sludge = 737.3 kg/d

Why the other options are there

F/M = 1,144 (basin volume omitted)
P_x = 1,135 kg/d (decay term ignored)

Reference: FE Reference Handbook — Environmental Engineering → Methanol Requirement for Biologically Treated Wastewater

Example 5

Activated sludge F/M ratio, aeration time and sludge production — Methanol Requirement for Biologically Treated Wastewater (5)

An activated sludge plant treats 4,517 m³/d with an influent BOD of 367 mg/L, effluent BOD 18 mg/L, MLSS 3,789 mg/L, and an aeration basin volume of 5,935 m³. With Y = 0.45 mg VSS/mg BOD, k_d = 0.08 d⁻¹ and an SRT of 15 days, compute the F/M ratio, hydraulic detention time and daily sludge production.

Given

Q = 4,517 m³/d
S₀ = 367 mg/L, S = 18 mg/L
X = 3,789 mg/L, V = 5,935 m³
Y = 0.45, k_d = 0.08 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
$F / M = 4517 (367) / (5935 \times 3789) = 0.074 d^{- 1}$
Formula
$τ = V / Q$
Substituting
$τ = 5935/4517 = 1.3139 d = 31.53 h$
Formula
$P_{x} = Y Q (S_{0} - S) / (1 + k_{d} \cdot S R T)$
Substituting
$P_{x} = 0.45 (4517) (367 - 18) / (1 + 0.08 \times 15) m g / d$
Evaluate
$P_{x} = 322.5 k g V S S / d a y$

Answer: F/M = 0.07 d⁻¹, τ = 31.5 h, sludge = 322.5 kg/d

Why the other options are there

F/M = 437.5 (basin volume omitted)
P_x = 709.4 kg/d (decay term ignored)

Reference: FE Reference Handbook — Environmental Engineering → Methanol Requirement for Biologically Treated Wastewater

Example 6

Activated sludge F/M ratio, aeration time and sludge production — Methanol Requirement for Biologically Treated Wastewater (6)

An activated sludge plant treats 8,047 m³/d with an influent BOD of 266 mg/L, effluent BOD 15 mg/L, MLSS 3,097 mg/L, and an aeration basin volume of 3,969 m³. With Y = 0.60 mg VSS/mg BOD, k_d = 0.06 d⁻¹ and an SRT of 19 days, compute the F/M ratio, hydraulic detention time and daily sludge production.

Given

Q = 8,047 m³/d
S₀ = 266 mg/L, S = 15 mg/L
X = 3,097 mg/L, V = 3,969 m³
Y = 0.60, k_d = 0.06 d⁻¹, SRT = 19 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
$F / M = 8047 (266) / (3969 \times 3097) = 0.174 d^{- 1}$
Formula
$τ = V / Q$
Substituting
$τ = 3969/8047 = 0.4932 d = 11.84 h$
Formula
$P_{x} = Y Q (S_{0} - S) / (1 + k_{d} \cdot S R T)$
Substituting
$P_{x} = 0.60 (8047) (266 - 15) / (1 + 0.06 \times 19) m g / d$
Evaluate
$P_{x} = 566.3 k g V S S / d a y$

Answer: F/M = 0.17 d⁻¹, τ = 11.8 h, sludge = 566.3 kg/d

Why the other options are there

F/M = 691.2 (basin volume omitted)
P_x = 1,212 kg/d (decay term ignored)

Reference: FE Reference Handbook — Environmental Engineering → Methanol Requirement for Biologically Treated Wastewater

Example 7

Activated sludge F/M ratio, aeration time and sludge production — Methanol Requirement for Biologically Treated Wastewater (7)

An activated sludge plant treats 2,951 m³/d with an influent BOD of 213 mg/L, effluent BOD 20 mg/L, MLSS 3,507 mg/L, and an aeration basin volume of 1,682 m³. With Y = 0.50 mg VSS/mg BOD, k_d = 0.04 d⁻¹ and an SRT of 16 days, compute the F/M ratio, hydraulic detention time and daily sludge production.

Given

Q = 2,951 m³/d
S₀ = 213 mg/L, S = 20 mg/L
X = 3,507 mg/L, V = 1,682 m³
Y = 0.50, k_d = 0.04 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
$F / M = 2951 (213) / (1682 \times 3507) = 0.107 d^{- 1}$
Formula
$τ = V / Q$
Substituting
$τ = 1682/2951 = 0.5700 d = 13.68 h$
Formula
$P_{x} = Y Q (S_{0} - S) / (1 + k_{d} \cdot S R T)$
Substituting
$P_{x} = 0.50 (2951) (213 - 20) / (1 + 0.04 \times 16) m g / d$
Evaluate
$P_{x} = 173.6 k g V S S / d a y$

Answer: F/M = 0.11 d⁻¹, τ = 13.7 h, sludge = 173.6 kg/d

Why the other options are there

F/M = 179.2 (basin volume omitted)
P_x = 284.8 kg/d (decay term ignored)

Reference: FE Reference Handbook — Environmental Engineering → Methanol Requirement for Biologically Treated Wastewater

Example 8

Activated sludge F/M ratio, aeration time and sludge production — Methanol Requirement for Biologically Treated Wastewater (8)

An activated sludge plant treats 24,421 m³/d with an influent BOD of 274 mg/L, effluent BOD 20 mg/L, MLSS 2,326 mg/L, and an aeration basin volume of 5,255 m³. With Y = 0.50 mg VSS/mg BOD, k_d = 0.08 d⁻¹ and an SRT of 18 days, compute the F/M ratio, hydraulic detention time and daily sludge production.

Given

Q = 24,421 m³/d
S₀ = 274 mg/L, S = 20 mg/L
X = 2,326 mg/L, V = 5,255 m³
Y = 0.50, k_d = 0.08 d⁻¹, SRT = 18 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
$F / M = 24421 (274) / (5255 \times 2326) = 0.547 d^{- 1}$
Formula
$τ = V / Q$
Substituting
$τ = 5255/24421 = 0.2152 d = 5.16 h$
Formula
$P_{x} = Y Q (S_{0} - S) / (1 + k_{d} \cdot S R T)$
Substituting
$P_{x} = 0.50 (24421) (274 - 20) / (1 + 0.08 \times 18) m g / d$
Evaluate
$P_{x} = 1, 271 k g V S S / d a y$

Answer: F/M = 0.55 d⁻¹, τ = 5.2 h, sludge = 1,271 kg/d

Why the other options are there

F/M = 2,877 (basin volume omitted)
P_x = 3,101 kg/d (decay term ignored)

Reference: FE Reference Handbook — Environmental Engineering → Methanol Requirement for Biologically Treated Wastewater

Example 9

Activated sludge F/M ratio, aeration time and sludge production — Methanol Requirement for Biologically Treated Wastewater (9)

An activated sludge plant treats 2,659 m³/d with an influent BOD of 205 mg/L, effluent BOD 7 mg/L, MLSS 3,439 mg/L, and an aeration basin volume of 1,927 m³. With Y = 0.60 mg VSS/mg BOD, k_d = 0.08 d⁻¹ and an SRT of 7 days, compute the F/M ratio, hydraulic detention time and daily sludge production.

Given

Q = 2,659 m³/d
S₀ = 205 mg/L, S = 7 mg/L
X = 3,439 mg/L, V = 1,927 m³
Y = 0.60, k_d = 0.08 d⁻¹, SRT = 7 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
$F / M = 2659 (205) / (1927 \times 3439) = 0.082 d^{- 1}$
Formula
$τ = V / Q$
Substituting
$τ = 1927/2659 = 0.7247 d = 17.39 h$
Formula
$P_{x} = Y Q (S_{0} - S) / (1 + k_{d} \cdot S R T)$
Substituting
$P_{x} = 0.60 (2659) (205 - 7) / (1 + 0.08 \times 7) m g / d$
Evaluate
$P_{x} = 202.5 k g V S S / d a y$

Answer: F/M = 0.08 d⁻¹, τ = 17.4 h, sludge = 202.5 kg/d

Why the other options are there

F/M = 158.5 (basin volume omitted)
P_x = 315.9 kg/d (decay term ignored)

Reference: FE Reference Handbook — Environmental Engineering → Methanol Requirement for Biologically Treated Wastewater

Example 10

Activated sludge F/M ratio, aeration time and sludge production — Methanol Requirement for Biologically Treated Wastewater (10)

An activated sludge plant treats 11,013 m³/d with an influent BOD of 220 mg/L, effluent BOD 15 mg/L, MLSS 2,939 mg/L, and an aeration basin volume of 822 m³. With Y = 0.45 mg VSS/mg BOD, k_d = 0.05 d⁻¹ and an SRT of 19 days, compute the F/M ratio, hydraulic detention time and daily sludge production.

Given

Q = 11,013 m³/d
S₀ = 220 mg/L, S = 15 mg/L
X = 2,939 mg/L, V = 822 m³
Y = 0.45, k_d = 0.05 d⁻¹, SRT = 19 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
$F / M = 11013 (220) / (822 \times 2939) = 1.003 d^{- 1}$
Formula
$τ = V / Q$
Substituting
$τ = 822/11013 = 0.0746 d = 1.79 h$
Formula
$P_{x} = Y Q (S_{0} - S) / (1 + k_{d} \cdot S R T)$
Substituting
$P_{x} = 0.45 (11013) (220 - 15) / (1 + 0.05 \times 19) m g / d$
Evaluate
$P_{x} = 521.0 k g V S S / d a y$

Answer: F/M = 1.00 d⁻¹, τ = 1.8 h, sludge = 521.0 kg/d

Why the other options are there

F/M = 824.4 (basin volume omitted)
P_x = 1,016 kg/d (decay term ignored)

Reference: FE Reference Handbook — Environmental Engineering → Methanol Requirement for Biologically Treated Wastewater

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

Methanol Requirement for Biologically Treated Wastewater contains 5 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.