Combustion

Environmental Engineering · FE Reference Handbook section

Environmental Engineering

12 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 Combustion 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 combustion describes physically and when it applies.
State every one of the 12 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. Combustion 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: combustion.
Capstone Studio instructional photograph

Environmental Engineering — Combustion: 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 12 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.

Combustible Substance Reaction Mols lb (kg)*
*Substitute the molecular weights in the reaction equation to secure lb (kg). The lb (kg) on each side of the equation must balance.
Hicks, Tyler G., Handbook of Energy Engineering Calculations, New York: McGraw-Hill, 2012.

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

Population projection and design water demand — Combustion

A city of 24,563 grows at 2.1% per year. Project the population in 29 years by both geometric and arithmetic growth, then compute the average and peak day water demand at 259 L/capita/day.

Given

P₀ = 24,563
i = 2.1%/yr
n = 29 yr
Per capita use = 259 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
$P = 24563 (1 + 0.021)^{2} 9 = 44, 877 p eo pl e$
Formula (arithmetic)
$P = P_{0} (1 + in)$
Substituting
$P = 24563 (1 + 0.021 \times 29) = 39, 522 p eo pl e$
Formula
$Q_{a} v g = P \times p er c a p i t a u se$
Substituting
$Q_{a} v g = 44, 877 (259) /1000 = 11, 623 m^{3} / d$
Peak day
$Q_{p} e ak = 1.9 \times 11, 623 = 22, 084 m^{3} / d$

Answer: P = 44,877 (geometric) vs 39,522 (arithmetic); Q_avg = 11,623 m³/d, Q_peak = 22,084 m³/d

Why the other options are there

14,959 (growth increment reported as population)
Q sized on the average day only

Reference: FE Reference Handbook — Environmental Engineering → Combustion

Example 2

Chronic daily intake, cancer risk and radioactive decay — Combustion

Drinking water contains 0.0640 mg/L of a carcinogen. An adult of 75 kg drinks 1.6 L/day. With a cancer slope factor of 1.77 (mg/kg·d)⁻¹, compute the chronic daily intake and the incremental lifetime cancer risk. Also determine the fraction of a radionuclide with a 44-year half-life remaining after 26 years.

Given

C = 0.0640 mg/L
IR = 1.6 L/d
BW = 75 kg
CSF = 1.77 (mg/kg·d)⁻¹
t½ = 44 yr, t = 26 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 = (0.0640 × 1.6)/75 = 1.365e-3 mg/kg·d
Formula
$R i s k = C D I \times C S F$
Substituting
$R i s k = 1.365 e - 3 (1.77) = 2.42 e - 3$
Interpretation — above the 10⁻⁶–10⁻⁴ risk range
Formula
$N / N_{0} = (½)^{(} t / t ½)$
Substituting
$(½)^{(} 26/44) = 66.392$

Answer: CDI = 1.37e-3 mg/kg·d, risk = 2.42e-3, 66.39% of the radionuclide remains

Why the other options are there

Risk = 0.11328 (body weight and intake ignored)
70.5% remaining (linear decay)

Reference: FE Reference Handbook — Environmental Engineering → Combustion

Example 3

Completely mixed stream blending — Combustion

A stream flowing 9.5 cfs at 18.0 mg/L receives a discharge of 6.0 cfs at 250.0 mg/L. Find the fully mixed concentration.

Given

Q₁ = 9.5 cfs, C₁ = 18.0 mg/L
Q₂ = 6.0 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
$9.5 (18.0) = 171.0; 6.0 (250.0) = 1, 500$
Total flow
$9.5 + 6.0 = 15.5 c f s$
Solve
$C = (171.0 + 1, 500) /15.5 = 107.8 m g / L$

Answer: C ≈ 107.8 mg/L

Why the other options are there

134.0 mg/L (unweighted average)
268.0 mg/L (concentrations added)

Reference: FE Reference Handbook — Environmental Engineering → Combustion

Example 4

BOD exerted after t days — Combustion

A wastewater has ultimate BOD L₀ = 353 mg/L and k = 0.29 /day (base e). How much BOD is exerted in 10 days?

Given

L₀ = 353 mg/L
k = 0.29 /day
t = 10 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
$k t = 0.29 (10) = 2.90$
Exponential
$e^{(} - 2.90) = 0.0550$
Substituting
$B O D_{t} = 353 (1 - 0.0550) = 333.6 m g / L$
Remaining
$L_{0} - B O D_{t} = 19.4 m g / L$

Answer: BOD_10 ≈ 333.6 mg/L

Why the other options are there

19 mg/L (remaining reported as exerted)
1,024 mg/L (linear decay assumed)

Reference: FE Reference Handbook — Environmental Engineering → Combustion

Example 5

Hydraulic detention time in a tank — Combustion

A treatment tank holds 155,031 gal and treats 4.7 MGD. Find the hydraulic detention time.

Given

V = 155,031 gal
Q = 4.7 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
$4.7 \times 1 0^{6} = 4, 700, 000 g a l / d a y$
Substituting
$θ = 155031/4, 700, 000 = 0.0330 d a y$
Convert
$θ = 0.0330 (24) = 0.79 h r$

Answer: θ ≈ 0.79 hr

Why the other options are there

0.033 hr (day/hour conversion missed)
30.32 hr (ratio inverted)

Reference: FE Reference Handbook — Environmental Engineering → Combustion

Example 6

Chemical feed rate from dosage — Combustion

A plant treating 7.1 MGD requires a 10.5 mg/L dose. What is the daily chemical demand in pounds?

Given

Q = 7.1 MGD
Dose = 10.5 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
$l b / d a y = 7.1 (10.5) (8.34)$
Evaluate — 621.7 lb/day
At 65% available strength — 956.5 lb/day of product

Answer: ≈ 621.7 lb/day

Why the other options are there

74.6 lb/day (8.34 omitted)
8.94 lb/day (factor divided)

Reference: FE Reference Handbook — Environmental Engineering → Combustion

Example 7

Converting ppm to mg/m³ — Combustion

A stack gas contains 42.0 ppm of a compound with molecular weight 44 g/mol at 25 °C and 1 atm. Express the concentration in mg/m³.

Given

C = 42.0 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
$m g / m^{3} = 42.0 (44) /24.45$
Evaluate — 75.58 mg/m³
Reverse check
$pp m = 75.58 (24.45) /44 = 42.0 ✓$

Answer: ≈ 75.6 mg/m³

Why the other options are there

23.3 mg/m³ (ratio inverted)
82.5 mg/m³ (0 °C molar volume used)

Reference: FE Reference Handbook — Environmental Engineering → Combustion

Example 8

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

A reactor of volume 1,883 m³ treats 1.30 m³/s carrying 386 mg/L of a contaminant that decays first-order with k = 0.15 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 = 1,883 m³
Q = 1.30 m³/s
C₀ = 386 mg/L
k = 0.15 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
$τ = 1883/1.30 = 1, 448 s = 0.402 h$
Formula (CSTR)
$C = C_{0} / (1 + k τ)$
Substituting
$C = 386/ (1 + 0.15 \times 0.402) = 364.0 m g / L$
Formula (PFR)
$C = C_{0} e^{(} - k τ)$
Substituting
$C = 386 e^{(} - 0.15 \times 0.402) = 363.4 m g / L$
Comparison — plug flow removes 5.9% versus 5.7% for the CSTR

Answer: τ = 0.40 h; C_CSTR = 364.0 mg/L, C_PFR = 363.4 mg/L

Why the other options are there

362.7 mg/L (linear decay assumed)
364.0 mg/L for both reactors

Reference: FE Reference Handbook — Environmental Engineering → Combustion

Example 9

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

An activated sludge plant treats 23,496 m³/d with an influent BOD of 319 mg/L, effluent BOD 6 mg/L, MLSS 3,256 mg/L, and an aeration basin volume of 3,479 m³. With Y = 0.60 mg VSS/mg BOD, k_d = 0.05 d⁻¹ and an SRT of 7 days, compute the F/M ratio, hydraulic detention time and daily sludge production.

Given

Q = 23,496 m³/d
S₀ = 319 mg/L, S = 6 mg/L
X = 3,256 mg/L, V = 3,479 m³
Y = 0.60, k_d = 0.05 d⁻¹, SRT = 7 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
$F / M = 23496 (319) / (3479 \times 3256) = 0.662 d^{- 1}$
Formula
$τ = V / Q$
Substituting
$τ = 3479/23496 = 0.1481 d = 3.55 h$
Formula
$P_{x} = Y Q (S_{0} - S) / (1 + k_{d} \cdot S R T)$
Substituting
$P_{x} = 0.60 (23496) (319 - 6) / (1 + 0.05 \times 7) m g / d$
Evaluate
$P_{x} = 3, 269 k g V S S / d a y$

Answer: F/M = 0.66 d⁻¹, τ = 3.6 h, sludge = 3,269 kg/d

Why the other options are there

F/M = 2,302 (basin volume omitted)
P_x = 4,413 kg/d (decay term ignored)

Reference: FE Reference Handbook — Environmental Engineering → Combustion

Example 10

Circular clarifier overflow rate, detention time and Stokes settling — Combustion

A circular clarifier 33 m in diameter and 3.0 m deep treats 3,807 m³/d. Compute the surface overflow rate, the detention time and the weir loading rate, then check whether a 40 µm particle (SG = 2.65) settles out by Stokes' law.

Given

Q = 3,807 m³/d
D = 33 m, depth = 3.0 m
d_p = 40 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
$A = (π /4) (33)^{2} = 855.3 m^{2}$
Formula
$S O R = Q / A$
Substituting — SOR = 3807/855.3 = 4.45 m³/m²·d
Formula — τ = V/Q = A·h/Q
Substituting
$τ = 2, 566/3807 = 16.18 h$
Formula
$w e i r l o a d in g = Q / (π D)$
Substituting — 3807/(π × 33) = 36.7 m³/m·d
Formula
$v_{s} = g (ρ_{s} - ρ) d^{2} / (18 μ)$
Substituting
$v_{s} = 9.81 (1650) (4.0 e - 5)^{2} / (18 \times 0.001) = 124.3 m / d$
Compare
$v_{s} = 124.3 m / d > S O R = 4.45 m / d \to p a r t i c l e i sr e m o v e d$

Answer: SOR = 4.5 m/d, τ = 16.2 h, weir loading = 37 m³/m·d, v_s = 124.3 m/d

Why the other options are there

SOR = 12.2 m/d (side area used)
v_s computed with the diameter unsquared

Reference: FE Reference Handbook — Environmental Engineering → Combustion

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

Combustion contains 12 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.

Combustion

Learning objectives

Lecture

Theory, developed

Handbook notes for this section

Core formulas for this FE topic

CarbontocarbonmonoxideC+1/2O2​=CO1+1/2=112+16=28

CarbontocarbondioxideC+O2​=CO2​1+1=112+16=28

CarbonmonoxidetocarbondioxideCO+1/2O2​=CO2​1+1/2=128+16=44

HydrogenH2​+1/2O2​=H2​O1+1/2=12+16=18

SulfurtosulfurdioxideS+O2​=SO2​1+1=132+32=64

SulfurtosulfurtrioxideS+3/2O2​=SO3​1+2/2=132+48=80

MethaneCH4​+2O2​=CO2​+2H2​O1+2=1+216+64=44+36

EthaneC2​H6​+7/2O2​=2CO2​+3H2​O1+7/2=2+330+112=88+54

PropaneC3​H8​+5O2​=3CO2​+4H2​O1+5=3+444+160=132+72

ButaneC4​H10+13/2O2​=4CO2​+5H2​O1+12/2=4+558+208=176+90

AcetyleneC2​H2​+5/2O2​=2CO2​+H2​O1+5/2=2+226+80=88+18

EthyleneC2​H4​+3O2​=2CO2​+2H2​O1+3=2+228+96=88+36

Worked exam-style examples

Self-check

Chapter summary

Common traps in this section