Bioconcentration Factor BCF

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 Bioconcentration Factor BCF 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 bioconcentration factor bcf 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. Bioconcentration Factor BCF 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: bioconcentration factor bcf.
Capstone Studio instructional photograph

Environmental Engineering — Bioconcentration Factor BCF: 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

BCF	Quantity produced by "BCF = Corg /C" — read its definition and unit from the handbook line directly above the equation.
Corg	Quantity produced by "Corg = equilibrium concentration in organism (mg/kg or ppm)" — read its definition and unit from the handbook line directly above the equation.
C	Quantity produced by "C = concentration in water (ppm)" — 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.

The amount of a chemical to accumulate in aquatic organisms.
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 — Bioconcentration Factor BCF

A stream flowing 27.0 cfs at 22.0 mg/L receives a discharge of 8.5 cfs at 200.0 mg/L. Find the fully mixed concentration.

Given

Q₁ = 27.0 cfs, C₁ = 22.0 mg/L
Q₂ = 8.5 cfs, C₂ = 200.0 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
$27.0 (22.0) = 594.0; 8.5 (200.0) = 1, 700$
Total flow
$27.0 + 8.5 = 35.5 c f s$
Solve
$C = (594.0 + 1, 700) /35.5 = 64.62 m g / L$

Answer: C ≈ 64.6 mg/L

Why the other options are there

111.0 mg/L (unweighted average)
222.0 mg/L (concentrations added)

Reference: FE Reference Handbook — Environmental Engineering → Bioconcentration Factor BCF

Example 2

Completely mixed stream blending — Bioconcentration Factor BCF (2)

A stream flowing 35.0 cfs at 27.5 mg/L receives a discharge of 16.0 cfs at 65 mg/L. Find the fully mixed concentration.

Given

Q₁ = 35.0 cfs, C₁ = 27.5 mg/L
Q₂ = 16.0 cfs, C₂ = 65 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
$35.0 (27.5) = 962.5; 16.0 (65) = 1, 040$
Total flow
$35.0 + 16.0 = 51.0 c f s$
Solve
$C = (962.5 + 1, 040) /51.0 = 39.26 m g / L$

Answer: C ≈ 39.3 mg/L

Why the other options are there

46.3 mg/L (unweighted average)
92.5 mg/L (concentrations added)

Reference: FE Reference Handbook — Environmental Engineering → Bioconcentration Factor BCF

Example 3

Completely mixed stream blending — Bioconcentration Factor BCF (3)

A stream flowing 22.0 cfs at 5.5 mg/L receives a discharge of 1.0 cfs at 280.0 mg/L. Find the fully mixed concentration.

Given

Q₁ = 22.0 cfs, C₁ = 5.5 mg/L
Q₂ = 1.0 cfs, C₂ = 280.0 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
$22.0 (5.5) = 121.0; 1.0 (280.0) = 280.0$
Total flow
$22.0 + 1.0 = 23.0 c f s$
Solve
$C = (121.0 + 280.0) /23.0 = 17.43 m g / L$

Answer: C ≈ 17.4 mg/L

Why the other options are there

142.8 mg/L (unweighted average)
285.5 mg/L (concentrations added)

Reference: FE Reference Handbook — Environmental Engineering → Bioconcentration Factor BCF

Example 4

Completely mixed stream blending — Bioconcentration Factor BCF (4)

A stream flowing 15.0 cfs at 11.0 mg/L receives a discharge of 3.0 cfs at 285.0 mg/L. Find the fully mixed concentration.

Given

Q₁ = 15.0 cfs, C₁ = 11.0 mg/L
Q₂ = 3.0 cfs, C₂ = 285.0 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
$15.0 (11.0) = 165.0; 3.0 (285.0) = 855.0$
Total flow
$15.0 + 3.0 = 18.0 c f s$
Solve
$C = (165.0 + 855.0) /18.0 = 56.67 m g / L$

Answer: C ≈ 56.7 mg/L

Why the other options are there

148.0 mg/L (unweighted average)
296.0 mg/L (concentrations added)

Reference: FE Reference Handbook — Environmental Engineering → Bioconcentration Factor BCF

Example 5

Completely mixed stream blending — Bioconcentration Factor BCF (5)

A stream flowing 2.5 cfs at 12.0 mg/L receives a discharge of 13.0 cfs at 250.0 mg/L. Find the fully mixed concentration.

Given

Q₁ = 2.5 cfs, C₁ = 12.0 mg/L
Q₂ = 13.0 cfs, C₂ = 250.0 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
$2.5 (12.0) = 30.0; 13.0 (250.0) = 3, 250$
Total flow
$2.5 + 13.0 = 15.5 c f s$
Solve
$C = (30.0 + 3, 250) /15.5 = 211.6 m g / L$

Answer: C ≈ 211.6 mg/L

Why the other options are there

131.0 mg/L (unweighted average)
262.0 mg/L (concentrations added)

Reference: FE Reference Handbook — Environmental Engineering → Bioconcentration Factor BCF

Example 6

Completely mixed stream blending — Bioconcentration Factor BCF (6)

A stream flowing 35.5 cfs at 1.5 mg/L receives a discharge of 18.0 cfs at 60 mg/L. Find the fully mixed concentration.

Given

Q₁ = 35.5 cfs, C₁ = 1.5 mg/L
Q₂ = 18.0 cfs, C₂ = 60 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
$35.5 (1.5) = 53.3; 18.0 (60) = 1, 080$
Total flow
$35.5 + 18.0 = 53.5 c f s$
Solve
$C = (53.3 + 1, 080) /53.5 = 21.18 m g / L$

Answer: C ≈ 21.2 mg/L

Why the other options are there

30.8 mg/L (unweighted average)
61.5 mg/L (concentrations added)

Reference: FE Reference Handbook — Environmental Engineering → Bioconcentration Factor BCF

Example 7

Completely mixed stream blending — Bioconcentration Factor BCF (7)

A stream flowing 29.0 cfs at 30.0 mg/L receives a discharge of 9.5 cfs at 175.0 mg/L. Find the fully mixed concentration.

Given

Q₁ = 29.0 cfs, C₁ = 30.0 mg/L
Q₂ = 9.5 cfs, C₂ = 175.0 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
$29.0 (30.0) = 870.0; 9.5 (175.0) = 1, 663$
Total flow
$29.0 + 9.5 = 38.5 c f s$
Solve
$C = (870.0 + 1, 663) /38.5 = 65.78 m g / L$

Answer: C ≈ 65.8 mg/L

Why the other options are there

102.5 mg/L (unweighted average)
205.0 mg/L (concentrations added)

Reference: FE Reference Handbook — Environmental Engineering → Bioconcentration Factor BCF

Example 8

Completely mixed stream blending — Bioconcentration Factor BCF (8)

A stream flowing 5.0 cfs at 12.5 mg/L receives a discharge of 14.5 cfs at 230.0 mg/L. Find the fully mixed concentration.

Given

Q₁ = 5.0 cfs, C₁ = 12.5 mg/L
Q₂ = 14.5 cfs, C₂ = 230.0 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
$5.0 (12.5) = 62.5; 14.5 (230.0) = 3, 335$
Total flow
$5.0 + 14.5 = 19.5 c f s$
Solve
$C = (62.5 + 3, 335) /19.5 = 174.2 m g / L$

Answer: C ≈ 174.2 mg/L

Why the other options are there

121.3 mg/L (unweighted average)
242.5 mg/L (concentrations added)

Reference: FE Reference Handbook — Environmental Engineering → Bioconcentration Factor BCF

Example 9

Completely mixed stream blending — Bioconcentration Factor BCF (9)

A stream flowing 11.5 cfs at 27.0 mg/L receives a discharge of 15.0 cfs at 60 mg/L. Find the fully mixed concentration.

Given

Q₁ = 11.5 cfs, C₁ = 27.0 mg/L
Q₂ = 15.0 cfs, C₂ = 60 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
$11.5 (27.0) = 310.5; 15.0 (60) = 900.0$
Total flow
$11.5 + 15.0 = 26.5 c f s$
Solve
$C = (310.5 + 900.0) /26.5 = 45.68 m g / L$

Answer: C ≈ 45.7 mg/L

Why the other options are there

43.5 mg/L (unweighted average)
87.0 mg/L (concentrations added)

Reference: FE Reference Handbook — Environmental Engineering → Bioconcentration Factor BCF

Example 10

Completely mixed stream blending — Bioconcentration Factor BCF (10)

A stream flowing 30.5 cfs at 19.0 mg/L receives a discharge of 18.0 cfs at 120.0 mg/L. Find the fully mixed concentration.

Given

Q₁ = 30.5 cfs, C₁ = 19.0 mg/L
Q₂ = 18.0 cfs, C₂ = 120.0 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
$30.5 (19.0) = 579.5; 18.0 (120.0) = 2, 160$
Total flow
$30.5 + 18.0 = 48.5 c f s$
Solve
$C = (579.5 + 2, 160) /48.5 = 56.48 m g / L$

Answer: C ≈ 56.5 mg/L

Why the other options are there

69.5 mg/L (unweighted average)
139.0 mg/L (concentrations added)

Reference: FE Reference Handbook — Environmental Engineering → Bioconcentration Factor BCF

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

Bioconcentration Factor BCF 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.