Phosphorus Removal Equations

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 Phosphorus Removal Equations 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 phosphorus removal equations 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. Phosphorus Removal Equations 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: phosphorus removal equations.
Capstone Studio instructional photograph

Environmental Engineering — Phosphorus Removal Equations: 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.

1. Ferric chloride
FeCl3 + PO43– → FePO4 (↓) + 3 Cl–
2. Ferrous chloride
3 FeCl2 + 2 PO43– → Fe3(PO4)2 (↓) + 6 Cl–
3. Aluminum sulfate (alum)
Al2 (SO4)3 • 14 H2O + 2 PO43– → 2 AlPO4 (↓) + 3 SO42– + 14 H2O
Common Radicals in Water
Molecular Molecular Equivalent
Formulas Weight # Equiv per Weight
mole
CO32– 60.0 2 30.0
CO2 44.0 2 22.0
Ca(OH)2 74.1 2 37.1
CaCO3 100.1 2 50.0
Ca(HCO 3)2 162.1 2 81.1
CaSO4 136.1 2 68.1
Ca2+ 40.1 2 20.0
H+ 1.0 1 1.0
HCO3– 61.0 1 61.0
Mg(HCO3)2 146.3 2 73.2
Mg(OH)2 58.3 2 29.2
MgSO4 120.4 2 60.2
Mg2+ 24.3 2 12.2
Na+ 23.0 1 23.0

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

First-order removal in a CSTR versus a plug-flow reactor — Phosphorus Removal Equations

A reactor of volume 1,708 m³ treats 0.70 m³/s carrying 289 mg/L of a contaminant that decays first-order with k = 0.55 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 = 1,708 m³
Q = 0.70 m³/s
C₀ = 289 mg/L
k = 0.55 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
$τ = 1708/0.70 = 2, 440 s = 0.678 h$
Formula (CSTR)
$C = C_{0} / (1 + k τ)$
Substituting
$C = 289/ (1 + 0.55 \times 0.678) = 210.5 m g / L$
Formula (PFR)
$C = C_{0} e^{(} - k τ)$
Substituting
$C = 289 e^{(} - 0.55 \times 0.678) = 199.1 m g / L$
Comparison — plug flow removes 31.1% versus 27.2% for the CSTR

Answer: τ = 0.68 h; C_CSTR = 210.5 mg/L, C_PFR = 199.1 mg/L

Why the other options are there

181.3 mg/L (linear decay assumed)
210.5 mg/L for both reactors

Reference: FE Reference Handbook — Environmental Engineering → Phosphorus Removal Equations

Example 2

Circular clarifier overflow rate, detention time and Stokes settling — Phosphorus Removal Equations

A circular clarifier 25 m in diameter and 4.5 m deep treats 20,515 m³/d. Compute the surface overflow rate, the detention time and the weir loading rate, then check whether a 50 µm particle (SG = 2.65) settles out by Stokes' law.

Given

Q = 20,515 m³/d
D = 25 m, depth = 4.5 m
d_p = 50 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
$A = (π /4) (25)^{2} = 490.9 m^{2}$
Formula
$S O R = Q / A$
Substituting — SOR = 20515/490.9 = 41.79 m³/m²·d
Formula — τ = V/Q = A·h/Q
Substituting
$τ = 2, 209/20515 = 2.58 h$
Formula
$w e i r l o a d in g = Q / (π D)$
Substituting — 20515/(π × 25) = 261.2 m³/m·d
Formula
$v_{s} = g (ρ_{s} - ρ) d^{2} / (18 μ)$
Substituting
$v_{s} = 9.81 (1650) (5.0 e - 5)^{2} / (18 \times 0.001) = 194.2 m / d$
Compare
$v_{s} = 194.2 m / d > S O R = 41.79 m / d \to p a r t i c l e i sr e m o v e d$

Answer: SOR = 41.8 m/d, τ = 2.6 h, weir loading = 261.2 m³/m·d, v_s = 194.2 m/d

Why the other options are there

SOR = 58.0 m/d (side area used)
v_s computed with the diameter unsquared

Reference: FE Reference Handbook — Environmental Engineering → Phosphorus Removal Equations

Example 3

Chlorine dose, CT value and daily chemical demand — Phosphorus Removal Equations

A water plant treating 35,715 m³/d applies a chlorine dose of 3.5 mg/L against a demand of 2.4 mg/L, with 45 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 = 35,715 m³/d
Dose = 3.5 mg/L
Demand = 2.4 mg/L
t = 45 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
$3.5 - 2.4 = 1.10 m g / L$
Formula
$C T = C \times t$
Substituting — CT = 1.10(45) = 49.5 mg·min/L
Formula
$ma ss = d ose \times Q$
Substituting
$3.5 g / m^{3} \times 35715 m^{3} / d \div 1000 = 125.0 k g / d a y$
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 = 1.10 mg/L, CT = 50 mg·min/L, 125.0 kg Cl₂/day, survivors 1.0e+3/100 mL

Why the other options are there

CT = 157.5 (dose used instead of residual)
125,003 kg/day (unit conversion missed)

Reference: FE Reference Handbook — Environmental Engineering → Phosphorus Removal Equations

Example 4

First-order removal in a CSTR versus a plug-flow reactor — Phosphorus Removal Equations (2)

A reactor of volume 1,514 m³ treats 1.35 m³/s carrying 316 mg/L of a contaminant that decays first-order with k = 0.25 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 = 1,514 m³
Q = 1.35 m³/s
C₀ = 316 mg/L
k = 0.25 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
$τ = 1514/1.35 = 1, 121 s = 0.312 h$
Formula (CSTR)
$C = C_{0} / (1 + k τ)$
Substituting
$C = 316/ (1 + 0.25 \times 0.312) = 293.2 m g / L$
Formula (PFR)
$C = C_{0} e^{(} - k τ)$
Substituting
$C = 316 e^{(} - 0.25 \times 0.312) = 292.3 m g / L$
Comparison — plug flow removes 7.5% versus 7.2% for the CSTR

Answer: τ = 0.31 h; C_CSTR = 293.2 mg/L, C_PFR = 292.3 mg/L

Why the other options are there

291.4 mg/L (linear decay assumed)
293.2 mg/L for both reactors

Reference: FE Reference Handbook — Environmental Engineering → Phosphorus Removal Equations

Example 5

Circular clarifier overflow rate, detention time and Stokes settling — Phosphorus Removal Equations (2)

A circular clarifier 29 m in diameter and 5.0 m deep treats 27,635 m³/d. Compute the surface overflow rate, the detention time and the weir loading rate, then check whether a 50 µm particle (SG = 2.65) settles out by Stokes' law.

Given

Q = 27,635 m³/d
D = 29 m, depth = 5.0 m
d_p = 50 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
$A = (π /4) (29)^{2} = 660.5 m^{2}$
Formula
$S O R = Q / A$
Substituting — SOR = 27635/660.5 = 41.84 m³/m²·d
Formula — τ = V/Q = A·h/Q
Substituting
$τ = 3, 303/27635 = 2.87 h$
Formula
$w e i r l o a d in g = Q / (π D)$
Substituting — 27635/(π × 29) = 303.3 m³/m·d
Formula
$v_{s} = g (ρ_{s} - ρ) d^{2} / (18 μ)$
Substituting
$v_{s} = 9.81 (1650) (5.0 e - 5)^{2} / (18 \times 0.001) = 194.2 m / d$
Compare
$v_{s} = 194.2 m / d > S O R = 41.84 m / d \to p a r t i c l e i sr e m o v e d$

Answer: SOR = 41.8 m/d, τ = 2.9 h, weir loading = 303.3 m³/m·d, v_s = 194.2 m/d

Why the other options are there

SOR = 60.7 m/d (side area used)
v_s computed with the diameter unsquared

Reference: FE Reference Handbook — Environmental Engineering → Phosphorus Removal Equations

Example 6

Chlorine dose, CT value and daily chemical demand — Phosphorus Removal Equations (2)

A water plant treating 35,971 m³/d applies a chlorine dose of 2.0 mg/L against a demand of 4.0 mg/L, with 117 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 = 35,971 m³/d
Dose = 2.0 mg/L
Demand = 4.0 mg/L
t = 117 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
$2.0 - 4.0 = 0.20 m g / L$
Formula
$C T = C \times t$
Substituting — CT = 0.20(117) = 23.4 mg·min/L
Formula
$ma ss = d ose \times Q$
Substituting
$2.0 g / m^{3} \times 35971 m^{3} / d \div 1000 = 71.94 k g / d a y$
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 = 0.20 mg/L, CT = 23 mg·min/L, 71.9 kg Cl₂/day, survivors 1.0e+2/100 mL

Why the other options are there

CT = 234.0 (dose used instead of residual)
71,942 kg/day (unit conversion missed)

Reference: FE Reference Handbook — Environmental Engineering → Phosphorus Removal Equations

Example 7

First-order removal in a CSTR versus a plug-flow reactor — Phosphorus Removal Equations (3)

A reactor of volume 4,335 m³ treats 0.15 m³/s carrying 384 mg/L of a contaminant that decays first-order with k = 0.45 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 = 4,335 m³
Q = 0.15 m³/s
C₀ = 384 mg/L
k = 0.45 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
$τ = 4335/0.15 = 28, 900 s = 8.028 h$
Formula (CSTR)
$C = C_{0} / (1 + k τ)$
Substituting
$C = 384/ (1 + 0.45 \times 8.028) = 83.25 m g / L$
Formula (PFR)
$C = C_{0} e^{(} - k τ)$
Substituting
$C = 384 e^{(} - 0.45 \times 8.028) = 10.36 m g / L$
Comparison — plug flow removes 97.3% versus 78.3% for the CSTR

Answer: τ = 8.03 h; C_CSTR = 83.3 mg/L, C_PFR = 10.4 mg/L

Why the other options are there

-1,003 mg/L (linear decay assumed)
83.3 mg/L for both reactors

Reference: FE Reference Handbook — Environmental Engineering → Phosphorus Removal Equations

Example 8

Circular clarifier overflow rate, detention time and Stokes settling — Phosphorus Removal Equations (3)

A circular clarifier 30 m in diameter and 5.0 m deep treats 15,805 m³/d. Compute the surface overflow rate, the detention time and the weir loading rate, then check whether a 50 µm particle (SG = 2.65) settles out by Stokes' law.

Given

Q = 15,805 m³/d
D = 30 m, depth = 5.0 m
d_p = 50 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
$A = (π /4) (30)^{2} = 706.9 m^{2}$
Formula
$S O R = Q / A$
Substituting — SOR = 15805/706.9 = 22.36 m³/m²·d
Formula — τ = V/Q = A·h/Q
Substituting
$τ = 3, 534/15805 = 5.37 h$
Formula
$w e i r l o a d in g = Q / (π D)$
Substituting — 15805/(π × 30) = 167.7 m³/m·d
Formula
$v_{s} = g (ρ_{s} - ρ) d^{2} / (18 μ)$
Substituting
$v_{s} = 9.81 (1650) (5.0 e - 5)^{2} / (18 \times 0.001) = 194.2 m / d$
Compare
$v_{s} = 194.2 m / d > S O R = 22.36 m / d \to p a r t i c l e i sr e m o v e d$

Answer: SOR = 22.4 m/d, τ = 5.4 h, weir loading = 167.7 m³/m·d, v_s = 194.2 m/d

Why the other options are there

SOR = 33.5 m/d (side area used)
v_s computed with the diameter unsquared

Reference: FE Reference Handbook — Environmental Engineering → Phosphorus Removal Equations

Example 9

Chlorine dose, CT value and daily chemical demand — Phosphorus Removal Equations (3)

A water plant treating 34,174 m³/d applies a chlorine dose of 4.0 mg/L against a demand of 3.2 mg/L, with 108 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 = 34,174 m³/d
Dose = 4.0 mg/L
Demand = 3.2 mg/L
t = 108 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
$4.0 - 3.2 = 0.80 m g / L$
Formula
$C T = C \times t$
Substituting — CT = 0.80(108) = 86.4 mg·min/L
Formula
$ma ss = d ose \times Q$
Substituting
$4.0 g / m^{3} \times 34174 m^{3} / d \div 1000 = 136.7 k g / d a y$
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 = 0.80 mg/L, CT = 86 mg·min/L, 136.7 kg Cl₂/day, survivors 3.2e+2/100 mL

Why the other options are there

CT = 432.0 (dose used instead of residual)
136,696 kg/day (unit conversion missed)

Reference: FE Reference Handbook — Environmental Engineering → Phosphorus Removal Equations

Example 10

First-order removal in a CSTR versus a plug-flow reactor — Phosphorus Removal Equations (4)

A reactor of volume 476 m³ treats 1.05 m³/s carrying 370 mg/L of a contaminant that decays first-order with k = 0.45 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 = 476 m³
Q = 1.05 m³/s
C₀ = 370 mg/L
k = 0.45 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
$τ = 476/1.05 = 453.3 s = 0.126 h$
Formula (CSTR)
$C = C_{0} / (1 + k τ)$
Substituting
$C = 370/ (1 + 0.45 \times 0.126) = 350.2 m g / L$
Formula (PFR)
$C = C_{0} e^{(} - k τ)$
Substituting
$C = 370 e^{(} - 0.45 \times 0.126) = 349.6 m g / L$
Comparison — plug flow removes 5.5% versus 5.4% for the CSTR

Answer: τ = 0.13 h; C_CSTR = 350.2 mg/L, C_PFR = 349.6 mg/L

Why the other options are there

349.0 mg/L (linear decay assumed)
350.2 mg/L for both reactors

Reference: FE Reference Handbook — Environmental Engineering → Phosphorus Removal Equations

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

Phosphorus Removal Equations 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.