Approximate Volume Flow Rate of Outdoor Air

Environmental Engineering · FE Reference Handbook section

Environmental Engineering

4 formulas

10 exam-style examples

~53 min

All Environmental Engineering lectures

Learning objectives

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

This chapter section covers Approximate Volume Flow Rate of Outdoor Air 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 approximate volume flow rate of outdoor air describes physically and when it applies.
State every one of the 4 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. Approximate Volume Flow Rate of Outdoor Air 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: approximate volume flow rate of outdoor air.
Capstone Studio instructional photograph

Environmental Engineering — Approximate Volume Flow Rate of Outdoor Air: 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 4 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

QOA	Quantity produced by "QOA = approximate volume flow rate of outdoor air (cfm)" — read its definition and unit from the handbook line directly above the equation.
n	Quantity produced by "n = number of people working in an office complex" — read its definition and unit from the handbook line directly above the equation.
Cindoors	Quantity produced by "Cindoors = measured concentration of tracer gas (e.g., CO2) in the space after a long period of time (e.g., 4 or more hours) of" — read its definition and unit from the handbook line directly above the equation.
COA	Quantity produced by "COA = concentration of the tracer gas (e.g., CO2) in the outdoor air (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.

13, 000 n
QOA . C
indoors - COA
where
human occupation (ppm)

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

Converting ppm to mg/m³ — Approximate Volume Flow Rate of Outdoor Air

A stack gas contains 19.0 ppm of a compound with molecular weight 28 g/mol at 25 °C and 1 atm. Express the concentration in mg/m³.

Given

C = 19.0 ppm
MW = 28 g/mol
Molar volume = 24.45 L/mol

Find

Concentration in mg/m³

Start with the thinking

ppm by volume needs the molar volume to become a mass concentration.
24.45 L/mol applies at 25 °C; use 22.4 at 0 °C.

Step-by-step solution

Conversion
$m g / m^{3} = pp m \times M W /24.45$
Substituting
$m g / m^{3} = 19.0 (28) /24.45$
Evaluate — 21.76 mg/m³
Reverse check
$pp m = 21.76 (24.45) /28 = 19.0 ✓$

Answer: ≈ 21.8 mg/m³

Why the other options are there

16.6 mg/m³ (ratio inverted)
23.8 mg/m³ (0 °C molar volume used)

Reference: FE Reference Handbook — Environmental Engineering → Approximate Volume Flow Rate of Outdoor Air

Example 2

Converting ppm to mg/m³ — Approximate Volume Flow Rate of Outdoor Air (2)

A stack gas contains 48.5 ppm of a compound with molecular weight 46 g/mol at 25 °C and 1 atm. Express the concentration in mg/m³.

Given

C = 48.5 ppm
MW = 46 g/mol
Molar volume = 24.45 L/mol

Find

Concentration in mg/m³

Start with the thinking

ppm by volume needs the molar volume to become a mass concentration.
24.45 L/mol applies at 25 °C; use 22.4 at 0 °C.

Step-by-step solution

Conversion
$m g / m^{3} = pp m \times M W /24.45$
Substituting
$m g / m^{3} = 48.5 (46) /24.45$
Evaluate — 91.25 mg/m³
Reverse check
$pp m = 91.25 (24.45) /46 = 48.5 ✓$

Answer: ≈ 91.2 mg/m³

Why the other options are there

25.8 mg/m³ (ratio inverted)
99.6 mg/m³ (0 °C molar volume used)

Reference: FE Reference Handbook — Environmental Engineering → Approximate Volume Flow Rate of Outdoor Air

Example 3

Converting ppm to mg/m³ — Approximate Volume Flow Rate of Outdoor Air (3)

A stack gas contains 19.0 ppm of a compound with molecular weight 44 g/mol at 25 °C and 1 atm. Express the concentration in mg/m³.

Given

C = 19.0 ppm
MW = 44 g/mol
Molar volume = 24.45 L/mol

Find

Concentration in mg/m³

Start with the thinking

ppm by volume needs the molar volume to become a mass concentration.
24.45 L/mol applies at 25 °C; use 22.4 at 0 °C.

Step-by-step solution

Conversion
$m g / m^{3} = pp m \times M W /24.45$
Substituting
$m g / m^{3} = 19.0 (44) /24.45$
Evaluate — 34.19 mg/m³
Reverse check
$pp m = 34.19 (24.45) /44 = 19.0 ✓$

Answer: ≈ 34.2 mg/m³

Why the other options are there

10.6 mg/m³ (ratio inverted)
37.3 mg/m³ (0 °C molar volume used)

Reference: FE Reference Handbook — Environmental Engineering → Approximate Volume Flow Rate of Outdoor Air

Example 4

Converting ppm to mg/m³ — Approximate Volume Flow Rate of Outdoor Air (4)

A stack gas contains 24.0 ppm of a compound with molecular weight 28 g/mol at 25 °C and 1 atm. Express the concentration in mg/m³.

Given

C = 24.0 ppm
MW = 28 g/mol
Molar volume = 24.45 L/mol

Find

Concentration in mg/m³

Start with the thinking

ppm by volume needs the molar volume to become a mass concentration.
24.45 L/mol applies at 25 °C; use 22.4 at 0 °C.

Step-by-step solution

Conversion
$m g / m^{3} = pp m \times M W /24.45$
Substituting
$m g / m^{3} = 24.0 (28) /24.45$
Evaluate — 27.48 mg/m³
Reverse check
$pp m = 27.48 (24.45) /28 = 24.0 ✓$

Answer: ≈ 27.5 mg/m³

Why the other options are there

21.0 mg/m³ (ratio inverted)
30.0 mg/m³ (0 °C molar volume used)

Reference: FE Reference Handbook — Environmental Engineering → Approximate Volume Flow Rate of Outdoor Air

Example 5

Converting ppm to mg/m³ — Approximate Volume Flow Rate of Outdoor Air (5)

A stack gas contains 42.5 ppm of a compound with molecular weight 28 g/mol at 25 °C and 1 atm. Express the concentration in mg/m³.

Given

C = 42.5 ppm
MW = 28 g/mol
Molar volume = 24.45 L/mol

Find

Concentration in mg/m³

Start with the thinking

ppm by volume needs the molar volume to become a mass concentration.
24.45 L/mol applies at 25 °C; use 22.4 at 0 °C.

Step-by-step solution

Conversion
$m g / m^{3} = pp m \times M W /24.45$
Substituting
$m g / m^{3} = 42.5 (28) /24.45$
Evaluate — 48.67 mg/m³
Reverse check
$pp m = 48.67 (24.45) /28 = 42.5 ✓$

Answer: ≈ 48.7 mg/m³

Why the other options are there

37.1 mg/m³ (ratio inverted)
53.1 mg/m³ (0 °C molar volume used)

Reference: FE Reference Handbook — Environmental Engineering → Approximate Volume Flow Rate of Outdoor Air

Example 6

Converting ppm to mg/m³ — Approximate Volume Flow Rate of Outdoor Air (6)

A stack gas contains 28.0 ppm of a compound with molecular weight 46 g/mol at 25 °C and 1 atm. Express the concentration in mg/m³.

Given

C = 28.0 ppm
MW = 46 g/mol
Molar volume = 24.45 L/mol

Find

Concentration in mg/m³

Start with the thinking

ppm by volume needs the molar volume to become a mass concentration.
24.45 L/mol applies at 25 °C; use 22.4 at 0 °C.

Step-by-step solution

Conversion
$m g / m^{3} = pp m \times M W /24.45$
Substituting
$m g / m^{3} = 28.0 (46) /24.45$
Evaluate — 52.68 mg/m³
Reverse check
$pp m = 52.68 (24.45) /46 = 28.0 ✓$

Answer: ≈ 52.7 mg/m³

Why the other options are there

14.9 mg/m³ (ratio inverted)
57.5 mg/m³ (0 °C molar volume used)

Reference: FE Reference Handbook — Environmental Engineering → Approximate Volume Flow Rate of Outdoor Air

Example 7

Converting ppm to mg/m³ — Approximate Volume Flow Rate of Outdoor Air (7)

A stack gas contains 44.5 ppm of a compound with molecular weight 28 g/mol at 25 °C and 1 atm. Express the concentration in mg/m³.

Given

C = 44.5 ppm
MW = 28 g/mol
Molar volume = 24.45 L/mol

Find

Concentration in mg/m³

Start with the thinking

ppm by volume needs the molar volume to become a mass concentration.
24.45 L/mol applies at 25 °C; use 22.4 at 0 °C.

Step-by-step solution

Conversion
$m g / m^{3} = pp m \times M W /24.45$
Substituting
$m g / m^{3} = 44.5 (28) /24.45$
Evaluate — 50.96 mg/m³
Reverse check
$pp m = 50.96 (24.45) /28 = 44.5 ✓$

Answer: ≈ 51.0 mg/m³

Why the other options are there

38.9 mg/m³ (ratio inverted)
55.6 mg/m³ (0 °C molar volume used)

Reference: FE Reference Handbook — Environmental Engineering → Approximate Volume Flow Rate of Outdoor Air

Example 8

Converting ppm to mg/m³ — Approximate Volume Flow Rate of Outdoor Air (8)

A stack gas contains 52.5 ppm of a compound with molecular weight 64 g/mol at 25 °C and 1 atm. Express the concentration in mg/m³.

Given

C = 52.5 ppm
MW = 64 g/mol
Molar volume = 24.45 L/mol

Find

Concentration in mg/m³

Start with the thinking

ppm by volume needs the molar volume to become a mass concentration.
24.45 L/mol applies at 25 °C; use 22.4 at 0 °C.

Step-by-step solution

Conversion
$m g / m^{3} = pp m \times M W /24.45$
Substituting
$m g / m^{3} = 52.5 (64) /24.45$
Evaluate — 137.4 mg/m³
Reverse check
$pp m = 137.4 (24.45) /64 = 52.5 ✓$

Answer: ≈ 137.4 mg/m³

Why the other options are there

20.1 mg/m³ (ratio inverted)
150.0 mg/m³ (0 °C molar volume used)

Reference: FE Reference Handbook — Environmental Engineering → Approximate Volume Flow Rate of Outdoor Air

Example 9

Converting ppm to mg/m³ — Approximate Volume Flow Rate of Outdoor Air (9)

A stack gas contains 57.0 ppm of a compound with molecular weight 44 g/mol at 25 °C and 1 atm. Express the concentration in mg/m³.

Given

C = 57.0 ppm
MW = 44 g/mol
Molar volume = 24.45 L/mol

Find

Concentration in mg/m³

Start with the thinking

ppm by volume needs the molar volume to become a mass concentration.
24.45 L/mol applies at 25 °C; use 22.4 at 0 °C.

Step-by-step solution

Conversion
$m g / m^{3} = pp m \times M W /24.45$
Substituting
$m g / m^{3} = 57.0 (44) /24.45$
Evaluate — 102.6 mg/m³
Reverse check
$pp m = 102.6 (24.45) /44 = 57.0 ✓$

Answer: ≈ 102.6 mg/m³

Why the other options are there

31.7 mg/m³ (ratio inverted)
112.0 mg/m³ (0 °C molar volume used)

Reference: FE Reference Handbook — Environmental Engineering → Approximate Volume Flow Rate of Outdoor Air

Example 10

Converting ppm to mg/m³ — Approximate Volume Flow Rate of Outdoor Air (10)

A stack gas contains 27.5 ppm of a compound with molecular weight 28 g/mol at 25 °C and 1 atm. Express the concentration in mg/m³.

Given

C = 27.5 ppm
MW = 28 g/mol
Molar volume = 24.45 L/mol

Find

Concentration in mg/m³

Start with the thinking

ppm by volume needs the molar volume to become a mass concentration.
24.45 L/mol applies at 25 °C; use 22.4 at 0 °C.

Step-by-step solution

Conversion
$m g / m^{3} = pp m \times M W /24.45$
Substituting
$m g / m^{3} = 27.5 (28) /24.45$
Evaluate — 31.49 mg/m³
Reverse check
$pp m = 31.49 (24.45) /28 = 27.5 ✓$

Answer: ≈ 31.5 mg/m³

Why the other options are there

24.0 mg/m³ (ratio inverted)
34.4 mg/m³ (0 °C molar volume used)

Reference: FE Reference Handbook — Environmental Engineering → Approximate Volume Flow Rate of Outdoor Air

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

Approximate Volume Flow Rate of Outdoor Air contains 4 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.