Organic Carbon Partition Coefficient Koc

Environmental Engineering · FE Reference Handbook section

Environmental Engineering

3 formulas

10 exam-style examples

~51 min

All Environmental Engineering lectures

Learning objectives

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

This chapter section covers Organic Carbon Partition Coefficient Koc 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 organic carbon partition coefficient koc describes physically and when it applies.
State every one of the 3 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. Organic Carbon Partition Coefficient Koc 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: organic carbon partition coefficient koc.
Capstone Studio instructional photograph

Environmental Engineering — Organic Carbon Partition Coefficient Koc: 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 3 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

Koc	Quantity produced by "Koc = Csoil /Cwater" — read its definition and unit from the handbook line directly above the equation.
Csoil	Quantity produced by "Csoil = concentration of chemical in organic carbon component of soil (µg adsorbed/kg organic C, or ppb)" — read its definition and unit from the handbook line directly above the equation.
Cwater	Quantity produced by "Cwater = concentration of chemical in water (ppb or µg/kg)" — 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
Davis, M.L., and D. Cornwell, Introduction to Environmental Engineering, 3rd ed., McGraw-Hill, 1998.

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 — Organic Carbon Partition Coefficient Koc

A stream flowing 39.0 cfs at 1.5 mg/L receives a discharge of 6.5 cfs at 50 mg/L. Find the fully mixed concentration.

Given

Q₁ = 39.0 cfs, C₁ = 1.5 mg/L
Q₂ = 6.5 cfs, C₂ = 50 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
$39.0 (1.5) = 58.5; 6.5 (50) = 325.0$
Total flow
$39.0 + 6.5 = 45.5 c f s$
Solve
$C = (58.5 + 325.0) /45.5 = 8.43 m g / L$

Answer: C ≈ 8.4 mg/L

Why the other options are there

25.8 mg/L (unweighted average)
51.5 mg/L (concentrations added)

Reference: FE Reference Handbook — Environmental Engineering → Organic Carbon Partition Coefficient Koc

Example 2

Circular clarifier overflow rate, detention time and Stokes settling — Organic Carbon Partition Coefficient Koc

A circular clarifier 22 m in diameter and 3.0 m deep treats 9,599 m³/d. Compute the surface overflow rate, the detention time and the weir loading rate, then check whether a 60 µm particle (SG = 2.65) settles out by Stokes' law.

Given

Q = 9,599 m³/d
D = 22 m, depth = 3.0 m
d_p = 60 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
$A = (π /4) (22)^{2} = 380.1 m^{2}$
Formula
$S O R = Q / A$
Substituting — SOR = 9599/380.1 = 25.25 m³/m²·d
Formula — τ = V/Q = A·h/Q
Substituting
$τ = 1, 140/9599 = 2.85 h$
Formula
$w e i r l o a d in g = Q / (π D)$
Substituting — 9599/(π × 22) = 138.9 m³/m·d
Formula
$v_{s} = g (ρ_{s} - ρ) d^{2} / (18 μ)$
Substituting
$v_{s} = 9.81 (1650) (6.0 e - 5)^{2} / (18 \times 0.001) = 279.7 m / d$
Compare
$v_{s} = 279.7 m / d > S O R = 25.25 m / d \to p a r t i c l e i sr e m o v e d$

Answer: SOR = 25.3 m/d, τ = 2.9 h, weir loading = 138.9 m³/m·d, v_s = 279.7 m/d

Why the other options are there

SOR = 46.3 m/d (side area used)
v_s computed with the diameter unsquared

Reference: FE Reference Handbook — Environmental Engineering → Organic Carbon Partition Coefficient Koc

Example 3

Series particulate control: overall collection efficiency and emission rate — Organic Carbon Partition Coefficient Koc

A stack gas stream of 49 m³/s carries 3 g/m³ of particulate. It passes a cyclone at 78% efficiency followed by a fabric filter at 96.0% efficiency. Compute the concentration after each device, the overall efficiency, and the emission rate in kg/h.

Given

Q = 49 m³/s
C_in = 3 g/m³
η₁ = 0.78
η₂ = 0.960

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
$C_{1} = 3 (1 - 0.78) = 0.660 g / m^{3}$
Formula
$C_{2} = C_{1} (1 - η_{2})$
Substituting
$C_{2} = 0.660 (1 - 0.960) = 0.0264 g / m^{3}$
Formula
$η_{o} v er a l l = 1 - (1 - η_{1}) (1 - η_{2})$
Substituting
$η = 1 - 0.220 \times 0.0400 = 0.99120 = 99.120$
Formula
$e mi ss i o n = C_{2} Q$
Substituting
$0.0264 g / m^{3} \times 49 m^{3} / s \times 3600 s / h \div 1000 = 4.657 k g / h$

Answer: C_out = 0.0264 g/m³, η = 99.12%, emission = 4.66 kg/h

Why the other options are there

η = 174.0% (efficiencies added)
529.2 kg/h (uncontrolled rate)

Reference: FE Reference Handbook — Environmental Engineering → Organic Carbon Partition Coefficient Koc

Example 4

Landfill volume required for a community's solid waste — Organic Carbon Partition Coefficient Koc

A community of 239,234 people generates 2.3 kg/person/day of MSW and diverts 40% through recycling. Compacted in place at 675 kg/m³ with 15% additional daily cover volume, find the airspace needed for 24 years.

Given

Population = 239,234
Generation = 2.3 kg/cap/d
Diversion = 0.40
Compacted density = 675 kg/m³
Cover = 15%, design life 24 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
$239234 (2.3) (1 - 0.40) = 330, 143 k g / d a y$
Formula
$V = ma ss / d e n s i t y$
Substituting
$V = 330, 143/675 = 489.1 m^{3} / d a y$
Annual with cover
$489.1 \times 365 \times 1.15 = 205, 300 m^{3} / y r$
Design life
$205, 300 \times 24 = 4, 927, 200 m^{3}$

Answer: ≈ 205,300 m³/yr, or 4,927,200 m³ over 24 years

Why the other options are there

7,140,869 m³ (diversion and cover ignored)
2,892,051,979 m³ (mass reported as volume)

Reference: FE Reference Handbook — Environmental Engineering → Organic Carbon Partition Coefficient Koc

Example 5

Chlorine dose, CT value and daily chemical demand — Organic Carbon Partition Coefficient Koc

A water plant treating 31,117 m³/d applies a chlorine dose of 2.5 mg/L against a demand of 1.0 mg/L, with 43 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 = 31,117 m³/d
Dose = 2.5 mg/L
Demand = 1.0 mg/L
t = 43 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(43) = 64.5 mg·min/L
Formula
$ma ss = d ose \times Q$
Substituting
$2.5 g / m^{3} \times 31117 m^{3} / d \div 1000 = 77.79 k g / d a y$
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 = 65 mg·min/L, 77.8 kg Cl₂/day, survivors 3.2e+2/100 mL

Why the other options are there

CT = 107.5 (dose used instead of residual)
77,793 kg/day (unit conversion missed)

Reference: FE Reference Handbook — Environmental Engineering → Organic Carbon Partition Coefficient Koc

Example 6

Population projection and design water demand — Organic Carbon Partition Coefficient Koc

A city of 97,096 grows at 3.0% per year. Project the population in 14 years by both geometric and arithmetic growth, then compute the average and peak day water demand at 378 L/capita/day.

Given

P₀ = 97,096
i = 3.0%/yr
n = 14 yr
Per capita use = 378 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
$P = 97096 (1 + 0.030)^{1} 4 = 146, 866 p eo pl e$
Formula (arithmetic)
$P = P_{0} (1 + in)$
Substituting
$P = 97096 (1 + 0.030 \times 14) = 137, 876 p eo pl e$
Formula
$Q_{a} v g = P \times p er c a p i t a u se$
Substituting
$Q_{a} v g = 146, 866 (378) /1000 = 55, 516 m^{3} / d$
Peak day
$Q_{p} e ak = 2.2 \times 55, 516 = 122, 134 m^{3} / d$

Answer: P = 146,866 (geometric) vs 137,876 (arithmetic); Q_avg = 55,516 m³/d, Q_peak = 122,134 m³/d

Why the other options are there

40,780 (growth increment reported as population)
Q sized on the average day only

Reference: FE Reference Handbook — Environmental Engineering → Organic Carbon Partition Coefficient Koc

Example 7

Chronic daily intake, cancer risk and radioactive decay — Organic Carbon Partition Coefficient Koc

Drinking water contains 0.0780 mg/L of a carcinogen. An adult of 82 kg drinks 1.7 L/day. With a cancer slope factor of 1.64 (mg/kg·d)⁻¹, compute the chronic daily intake and the incremental lifetime cancer risk. Also determine the fraction of a radionuclide with a 47-year half-life remaining after 75 years.

Given

C = 0.0780 mg/L
IR = 1.7 L/d
BW = 82 kg
CSF = 1.64 (mg/kg·d)⁻¹
t½ = 47 yr, t = 75 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 = (0.0780 × 1.7)/82 = 1.617e-3 mg/kg·d
Formula
$R i s k = C D I \times C S F$
Substituting
$R i s k = 1.617 e - 3 (1.64) = 2.65 e - 3$
Interpretation — above the 10⁻⁶–10⁻⁴ risk range
Formula
$N / N_{0} = (½)^{(} t / t ½)$
Substituting
$(½)^{(} 75/47) = 33.085$

Answer: CDI = 1.62e-3 mg/kg·d, risk = 2.65e-3, 33.09% of the radionuclide remains

Why the other options are there

Risk = 0.12792 (body weight and intake ignored)
20.2% remaining (linear decay)

Reference: FE Reference Handbook — Environmental Engineering → Organic Carbon Partition Coefficient Koc

Example 8

BOD exerted after t days — Organic Carbon Partition Coefficient Koc

A wastewater has ultimate BOD L₀ = 156 mg/L and k = 0.28 /day (base e). How much BOD is exerted in 5 days?

Given

L₀ = 156 mg/L
k = 0.28 /day
t = 5 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
$k t = 0.28 (5) = 1.40$
Exponential
$e^{(} - 1.40) = 0.2466$
Substituting
$B O D_{t} = 156 (1 - 0.2466) = 117.5 m g / L$
Remaining
$L_{0} - B O D_{t} = 38.5 m g / L$

Answer: BOD_5 ≈ 117.5 mg/L

Why the other options are there

38 mg/L (remaining reported as exerted)
218.4 mg/L (linear decay assumed)

Reference: FE Reference Handbook — Environmental Engineering → Organic Carbon Partition Coefficient Koc

Example 9

Hydraulic detention time in a tank — Organic Carbon Partition Coefficient Koc

A treatment tank holds 34,239 gal and treats 5.9 MGD. Find the hydraulic detention time.

Given

V = 34,239 gal
Q = 5.9 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
$5.9 \times 1 0^{6} = 5, 900, 000 g a l / d a y$
Substituting
$θ = 34239/5, 900, 000 = 0.0058 d a y$
Convert
$θ = 0.0058 (24) = 0.14 h r$

Answer: θ ≈ 0.14 hr

Why the other options are there

0.006 hr (day/hour conversion missed)
172.3 hr (ratio inverted)

Reference: FE Reference Handbook — Environmental Engineering → Organic Carbon Partition Coefficient Koc

Example 10

Chemical feed rate from dosage — Organic Carbon Partition Coefficient Koc

A plant treating 3.2 MGD requires a 10.0 mg/L dose. What is the daily chemical demand in pounds?

Given

Q = 3.2 MGD
Dose = 10.0 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
$l b / d a y = 3.2 (10.0) (8.34)$
Evaluate — 266.9 lb/day
At 65% available strength — 410.6 lb/day of product

Answer: ≈ 266.9 lb/day

Why the other options are there

32.0 lb/day (8.34 omitted)
3.84 lb/day (factor divided)

Reference: FE Reference Handbook — Environmental Engineering → Organic Carbon Partition Coefficient Koc

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

Organic Carbon Partition Coefficient Koc contains 3 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.