Mass Calculations

Environmental Engineering · FE Reference Handbook section

Environmental Engineering

18 formulas

10 exam-style examples

~60 min

All Environmental Engineering lectures

Learning objectives

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

This chapter section covers Mass Calculations 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 mass calculations describes physically and when it applies.
State every one of the 18 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. Mass Calculations 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: mass calculations.
Capstone Studio instructional photograph

Environmental Engineering — Mass Calculations: 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 18 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

M	Quantity produced by "M = CQ = CV" — read its definition and unit from the handbook line directly above the equation.
Min	Quantity produced by "Min = mass in" — read its definition and unit from the handbook line directly above the equation.
Mout	Quantity produced by "Mout = mass out" — read its definition and unit from the handbook line directly above the equation.
r	Quantity produced by "r = reaction rate = kCn" — read its definition and unit from the handbook line directly above the equation.
n	Quantity produced by "n = order of reaction" — read its definition and unit from the handbook line directly above the equation.
C	Quantity produced by "C = concentration (mass/volume)" — read its definition and unit from the handbook line directly above the equation.
Q	Quantity produced by "Q = flowrate" — read its definition and unit from the handbook line directly above the equation.
V	Quantity produced by "V = volume" — read its definition and unit from the handbook line directly above the equation.
v	Quantity produced by "v = velocity" — read its definition and unit from the handbook line directly above the equation.
A	Quantity produced by "A = cross-sectional area of flow" — read its definition and unit from the handbook line directly above the equation.
M (lb/day)	Quantity produced by "M (lb/day) = C (mg/L) × Q (MGD) × 8.34 [lb-L/(mg-MG)]" — read its definition and unit from the handbook line directly above the equation.
MGD	Quantity produced by "MGD = million gallons per day" — 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.

dM dM dM
where
k -
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

First-order removal in a CSTR versus a plug-flow reactor — Mass Calculations

A reactor of volume 4,586 m³ treats 0.25 m³/s carrying 112 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 = 4,586 m³
Q = 0.25 m³/s
C₀ = 112 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
$τ = 4586/0.25 = 18, 344 s = 5.096 h$
Formula (CSTR)
$C = C_{0} / (1 + k τ)$
Substituting
$C = 112/ (1 + 0.45 \times 5.096) = 34.01 m g / L$
Formula (PFR)
$C = C_{0} e^{(} - k τ)$
Substituting
$C = 112 e^{(} - 0.45 \times 5.096) = 11.31 m g / L$
Comparison — plug flow removes 89.9% versus 69.6% for the CSTR

Answer: τ = 5.10 h; C_CSTR = 34.0 mg/L, C_PFR = 11.3 mg/L

Why the other options are there

-144.8 mg/L (linear decay assumed)
34.0 mg/L for both reactors

Reference: FE Reference Handbook — Environmental Engineering → Mass Calculations

Example 2

First-order removal in a CSTR versus a plug-flow reactor — Mass Calculations (2)

A reactor of volume 4,765 m³ treats 0.45 m³/s carrying 173 mg/L of a contaminant that decays first-order with k = 0.20 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 = 4,765 m³
Q = 0.45 m³/s
C₀ = 173 mg/L
k = 0.20 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
$τ = 4765/0.45 = 10, 589 s = 2.941 h$
Formula (CSTR)
$C = C_{0} / (1 + k τ)$
Substituting
$C = 173/ (1 + 0.20 \times 2.941) = 108.9 m g / L$
Formula (PFR)
$C = C_{0} e^{(} - k τ)$
Substituting
$C = 173 e^{(} - 0.20 \times 2.941) = 96.06 m g / L$
Comparison — plug flow removes 44.5% versus 37.0% for the CSTR

Answer: τ = 2.94 h; C_CSTR = 108.9 mg/L, C_PFR = 96.1 mg/L

Why the other options are there

71.2 mg/L (linear decay assumed)
108.9 mg/L for both reactors

Reference: FE Reference Handbook — Environmental Engineering → Mass Calculations

Example 3

First-order removal in a CSTR versus a plug-flow reactor — Mass Calculations (3)

A reactor of volume 4,623 m³ treats 1.90 m³/s carrying 67 mg/L of a contaminant that decays first-order with k = 0.50 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 = 4,623 m³
Q = 1.90 m³/s
C₀ = 67 mg/L
k = 0.50 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
$τ = 4623/1.90 = 2, 433 s = 0.676 h$
Formula (CSTR)
$C = C_{0} / (1 + k τ)$
Substituting
$C = 67/ (1 + 0.50 \times 0.676) = 50.08 m g / L$
Formula (PFR)
$C = C_{0} e^{(} - k τ)$
Substituting
$C = 67 e^{(} - 0.50 \times 0.676) = 47.79 m g / L$
Comparison — plug flow removes 28.7% versus 25.3% for the CSTR

Answer: τ = 0.68 h; C_CSTR = 50.1 mg/L, C_PFR = 47.8 mg/L

Why the other options are there

44.4 mg/L (linear decay assumed)
50.1 mg/L for both reactors

Reference: FE Reference Handbook — Environmental Engineering → Mass Calculations

Example 4

First-order removal in a CSTR versus a plug-flow reactor — Mass Calculations (4)

A reactor of volume 3,194 m³ treats 0.40 m³/s carrying 316 mg/L of a contaminant that decays first-order with k = 0.35 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 = 3,194 m³
Q = 0.40 m³/s
C₀ = 316 mg/L
k = 0.35 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
$τ = 3194/0.40 = 7, 985 s = 2.218 h$
Formula (CSTR)
$C = C_{0} / (1 + k τ)$
Substituting
$C = 316/ (1 + 0.35 \times 2.218) = 177.9 m g / L$
Formula (PFR)
$C = C_{0} e^{(} - k τ)$
Substituting
$C = 316 e^{(} - 0.35 \times 2.218) = 145.4 m g / L$
Comparison — plug flow removes 54.0% versus 43.7% for the CSTR

Answer: τ = 2.22 h; C_CSTR = 177.9 mg/L, C_PFR = 145.4 mg/L

Why the other options are there

70.7 mg/L (linear decay assumed)
177.9 mg/L for both reactors

Reference: FE Reference Handbook — Environmental Engineering → Mass Calculations

Example 5

Completely mixed stream blending — Mass Calculations

A stream flowing 14.0 cfs at 29.5 mg/L receives a discharge of 19.5 cfs at 250.0 mg/L. Find the fully mixed concentration.

Given

Q₁ = 14.0 cfs, C₁ = 29.5 mg/L
Q₂ = 19.5 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
$14.0 (29.5) = 413.0; 19.5 (250.0) = 4, 875$
Total flow
$14.0 + 19.5 = 33.5 c f s$
Solve
$C = (413.0 + 4, 875) /33.5 = 157.9 m g / L$

Answer: C ≈ 157.9 mg/L

Why the other options are there

139.8 mg/L (unweighted average)
279.5 mg/L (concentrations added)

Reference: FE Reference Handbook — Environmental Engineering → Mass Calculations

Example 6

Converting ppm to mg/m³ — Mass Calculations

A stack gas contains 58.5 ppm of a compound with molecular weight 46 g/mol at 25 °C and 1 atm. Express the concentration in mg/m³.

Given

C = 58.5 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
$m g / m^{3} = 58.5 (46) /24.45$
Evaluate — 110.1 mg/m³
Reverse check
$pp m = 110.1 (24.45) /46 = 58.5 ✓$

Answer: ≈ 110.1 mg/m³

Why the other options are there

31.1 mg/m³ (ratio inverted)
120.1 mg/m³ (0 °C molar volume used)

Reference: FE Reference Handbook — Environmental Engineering → Mass Calculations

Example 7

Activated sludge F/M ratio, aeration time and sludge production — Mass Calculations

An activated sludge plant treats 19,680 m³/d with an influent BOD of 372 mg/L, effluent BOD 17 mg/L, MLSS 3,315 mg/L, and an aeration basin volume of 3,238 m³. With Y = 0.45 mg VSS/mg BOD, k_d = 0.04 d⁻¹ and an SRT of 14 days, compute the F/M ratio, hydraulic detention time and daily sludge production.

Given

Q = 19,680 m³/d
S₀ = 372 mg/L, S = 17 mg/L
X = 3,315 mg/L, V = 3,238 m³
Y = 0.45, k_d = 0.04 d⁻¹, SRT = 14 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
$F / M = 19680 (372) / (3238 \times 3315) = 0.682 d^{- 1}$
Formula
$τ = V / Q$
Substituting
$τ = 3238/19680 = 0.1645 d = 3.95 h$
Formula
$P_{x} = Y Q (S_{0} - S) / (1 + k_{d} \cdot S R T)$
Substituting
$P_{x} = 0.45 (19680) (372 - 17) / (1 + 0.04 \times 14) m g / d$
Evaluate
$P_{x} = 2, 015 k g V S S / d a y$

Answer: F/M = 0.68 d⁻¹, τ = 3.9 h, sludge = 2,015 kg/d

Why the other options are there

F/M = 2,208 (basin volume omitted)
P_x = 3,144 kg/d (decay term ignored)

Reference: FE Reference Handbook — Environmental Engineering → Mass Calculations

Example 8

Circular clarifier overflow rate, detention time and Stokes settling — Mass Calculations

A circular clarifier 26 m in diameter and 4.5 m deep treats 11,213 m³/d. Compute the surface overflow rate, the detention time and the weir loading rate, then check whether a 80 µm particle (SG = 2.65) settles out by Stokes' law.

Given

Q = 11,213 m³/d
D = 26 m, depth = 4.5 m
d_p = 80 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
$A = (π /4) (26)^{2} = 530.9 m^{2}$
Formula
$S O R = Q / A$
Substituting — SOR = 11213/530.9 = 21.12 m³/m²·d
Formula — τ = V/Q = A·h/Q
Substituting
$τ = 2, 389/11213 = 5.11 h$
Formula
$w e i r l o a d in g = Q / (π D)$
Substituting — 11213/(π × 26) = 137.3 m³/m·d
Formula
$v_{s} = g (ρ_{s} - ρ) d^{2} / (18 μ)$
Substituting
$v_{s} = 9.81 (1650) (8.0 e - 5)^{2} / (18 \times 0.001) = 497.2 m / d$
Compare
$v_{s} = 497.2 m / d > S O R = 21.12 m / d \to p a r t i c l e i sr e m o v e d$

Answer: SOR = 21.1 m/d, τ = 5.1 h, weir loading = 137.3 m³/m·d, v_s = 497.2 m/d

Why the other options are there

SOR = 30.5 m/d (side area used)
v_s computed with the diameter unsquared

Reference: FE Reference Handbook — Environmental Engineering → Mass Calculations

Example 9

Series particulate control: overall collection efficiency and emission rate — Mass Calculations

A stack gas stream of 46 m³/s carries 12 g/m³ of particulate. It passes a cyclone at 78% efficiency followed by a fabric filter at 92.5% efficiency. Compute the concentration after each device, the overall efficiency, and the emission rate in kg/h.

Given

Q = 46 m³/s
C_in = 12 g/m³
η₁ = 0.78
η₂ = 0.925

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
$C_{1} = 12 (1 - 0.78) = 2.640 g / m^{3}$
Formula
$C_{2} = C_{1} (1 - η_{2})$
Substituting
$C_{2} = 2.640 (1 - 0.925) = 0.1980 g / m^{3}$
Formula
$η_{o} v er a l l = 1 - (1 - η_{1}) (1 - η_{2})$
Substituting
$η = 1 - 0.220 \times 0.0750 = 0.98350 = 98.350$
Formula
$e mi ss i o n = C_{2} Q$
Substituting
$0.1980 g / m^{3} \times 46 m^{3} / s \times 3600 s / h \div 1000 = 32.789 k g / h$

Answer: C_out = 0.1980 g/m³, η = 98.35%, emission = 32.79 kg/h

Why the other options are there

η = 170.5% (efficiencies added)
1,987 kg/h (uncontrolled rate)

Reference: FE Reference Handbook — Environmental Engineering → Mass Calculations

Example 10

Landfill volume required for a community's solid waste — Mass Calculations

A community of 56,897 people generates 3.0 kg/person/day of MSW and diverts 45% through recycling. Compacted in place at 564 kg/m³ with 25% additional daily cover volume, find the airspace needed for 27 years.

Given

Population = 56,897
Generation = 3.0 kg/cap/d
Diversion = 0.45
Compacted density = 564 kg/m³
Cover = 25%, design life 27 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
$56897 (3.0) (1 - 0.45) = 93, 880 k g / d a y$
Formula
$V = ma ss / d e n s i t y$
Substituting
$V = 93, 880/564 = 166.5 m^{3} / d a y$
Annual with cover
$166.5 \times 365 \times 1.25 = 75, 945 m^{3} / y r$
Design life
$75, 945 \times 27 = 2, 050, 505 m^{3}$

Answer: ≈ 75,945 m³/yr, or 2,050,505 m³ over 27 years

Why the other options are there

2,982,553 m³ (diversion and cover ignored)
925,187,893 m³ (mass reported as volume)

Reference: FE Reference Handbook — Environmental Engineering → Mass Calculations

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

Mass Calculations contains 18 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.