Percent Growth

Environmental Engineering · FE Reference Handbook section

Environmental Engineering

5 formulas

10 exam-style examples

~55 min

All Environmental Engineering lectures

Learning objectives

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

This chapter section covers Percent Growth 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 percent growth describes physically and when it applies.
State every one of the 5 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. Percent Growth 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: percent growth.
Capstone Studio instructional photograph

Environmental Engineering — Percent Growth: 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 5 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

Pt	Quantity produced by "Pt = P0(1 + k)n" — read its definition and unit from the handbook line directly above the equation.
P0	Quantity produced by "P0 = population at time zero" — read its definition and unit from the handbook line directly above the equation.
k	Quantity produced by "k = growth rate" — read its definition and unit from the handbook line directly above the equation.
n	Quantity produced by "n = number of periods" — 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

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 — Percent Growth

A city of 103,281 grows at 1.5% per year. Project the population in 14 years by both geometric and arithmetic growth, then compute the average and peak day water demand at 206 L/capita/day.

Given

P₀ = 103,281
i = 1.5%/yr
n = 14 yr
Per capita use = 206 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
$P = 103281 (1 + 0.015)^{1} 4 = 127, 217 p eo pl e$
Formula (arithmetic)
$P = P_{0} (1 + in)$
Substituting
$P = 103281 (1 + 0.015 \times 14) = 124, 970 p eo pl e$
Formula
$Q_{a} v g = P \times p er c a p i t a u se$
Substituting
$Q_{a} v g = 127, 217 (206) /1000 = 26, 207 m^{3} / d$
Peak day
$Q_{p} e ak = 2.7 \times 26, 207 = 70, 758 m^{3} / d$

Answer: P = 127,217 (geometric) vs 124,970 (arithmetic); Q_avg = 26,207 m³/d, Q_peak = 70,758 m³/d

Why the other options are there

21,689 (growth increment reported as population)
Q sized on the average day only

Reference: FE Reference Handbook — Environmental Engineering → Percent Growth

Example 2

Population projection and design water demand — Percent Growth (2)

A city of 120,255 grows at 2.3% per year. Project the population in 19 years by both geometric and arithmetic growth, then compute the average and peak day water demand at 487 L/capita/day.

Given

P₀ = 120,255
i = 2.3%/yr
n = 19 yr
Per capita use = 487 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
$P = 120255 (1 + 0.023)^{1} 9 = 185, 242 p eo pl e$
Formula (arithmetic)
$P = P_{0} (1 + in)$
Substituting
$P = 120255 (1 + 0.023 \times 19) = 172, 806 p eo pl e$
Formula
$Q_{a} v g = P \times p er c a p i t a u se$
Substituting
$Q_{a} v g = 185, 242 (487) /1000 = 90, 213 m^{3} / d$
Peak day
$Q_{p} e ak = 2.4 \times 90, 213 = 216, 511 m^{3} / d$

Answer: P = 185,242 (geometric) vs 172,806 (arithmetic); Q_avg = 90,213 m³/d, Q_peak = 216,511 m³/d

Why the other options are there

52,551 (growth increment reported as population)
Q sized on the average day only

Reference: FE Reference Handbook — Environmental Engineering → Percent Growth

Example 3

Population projection and design water demand — Percent Growth (3)

A city of 122,118 grows at 1.2% per year. Project the population in 21 years by both geometric and arithmetic growth, then compute the average and peak day water demand at 499 L/capita/day.

Given

P₀ = 122,118
i = 1.2%/yr
n = 21 yr
Per capita use = 499 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
$P = 122118 (1 + 0.012)^{2} 1 = 156, 881 p eo pl e$
Formula (arithmetic)
$P = P_{0} (1 + in)$
Substituting
$P = 122118 (1 + 0.012 \times 21) = 152, 892 p eo pl e$
Formula
$Q_{a} v g = P \times p er c a p i t a u se$
Substituting
$Q_{a} v g = 156, 881 (499) /1000 = 78, 284 m^{3} / d$
Peak day
$Q_{p} e ak = 2.9 \times 78, 284 = 227, 023 m^{3} / d$

Answer: P = 156,881 (geometric) vs 152,892 (arithmetic); Q_avg = 78,284 m³/d, Q_peak = 227,023 m³/d

Why the other options are there

30,774 (growth increment reported as population)
Q sized on the average day only

Reference: FE Reference Handbook — Environmental Engineering → Percent Growth

Example 4

Population projection and design water demand — Percent Growth (4)

A city of 43,420 grows at 1.3% per year. Project the population in 10 years by both geometric and arithmetic growth, then compute the average and peak day water demand at 463 L/capita/day.

Given

P₀ = 43,420
i = 1.3%/yr
n = 10 yr
Per capita use = 463 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
$P = 43420 (1 + 0.013)^{1} 0 = 49, 407 p eo pl e$
Formula (arithmetic)
$P = P_{0} (1 + in)$
Substituting
$P = 43420 (1 + 0.013 \times 10) = 49, 065 p eo pl e$
Formula
$Q_{a} v g = P \times p er c a p i t a u se$
Substituting
$Q_{a} v g = 49, 407 (463) /1000 = 22, 875 m^{3} / d$
Peak day
$Q_{p} e ak = 2.3 \times 22, 875 = 52, 613 m^{3} / d$

Answer: P = 49,407 (geometric) vs 49,065 (arithmetic); Q_avg = 22,875 m³/d, Q_peak = 52,613 m³/d

Why the other options are there

5,645 (growth increment reported as population)
Q sized on the average day only

Reference: FE Reference Handbook — Environmental Engineering → Percent Growth

Example 5

Population projection and design water demand — Percent Growth (5)

A city of 49,675 grows at 2.9% per year. Project the population in 16 years by both geometric and arithmetic growth, then compute the average and peak day water demand at 250 L/capita/day.

Given

P₀ = 49,675
i = 2.9%/yr
n = 16 yr
Per capita use = 250 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
$P = 49675 (1 + 0.029)^{1} 6 = 78, 484 p eo pl e$
Formula (arithmetic)
$P = P_{0} (1 + in)$
Substituting
$P = 49675 (1 + 0.029 \times 16) = 72, 724 p eo pl e$
Formula
$Q_{a} v g = P \times p er c a p i t a u se$
Substituting
$Q_{a} v g = 78, 484 (250) /1000 = 19, 621 m^{3} / d$
Peak day
$Q_{p} e ak = 2.1 \times 19, 621 = 41, 204 m^{3} / d$

Answer: P = 78,484 (geometric) vs 72,724 (arithmetic); Q_avg = 19,621 m³/d, Q_peak = 41,204 m³/d

Why the other options are there

23,049 (growth increment reported as population)
Q sized on the average day only

Reference: FE Reference Handbook — Environmental Engineering → Percent Growth

Example 6

Population projection and design water demand — Percent Growth (6)

A city of 71,715 grows at 0.9% per year. Project the population in 24 years by both geometric and arithmetic growth, then compute the average and peak day water demand at 247 L/capita/day.

Given

P₀ = 71,715
i = 0.9%/yr
n = 24 yr
Per capita use = 247 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
$P = 71715 (1 + 0.009)^{2} 4 = 88, 920 p eo pl e$
Formula (arithmetic)
$P = P_{0} (1 + in)$
Substituting
$P = 71715 (1 + 0.009 \times 24) = 87, 205 p eo pl e$
Formula
$Q_{a} v g = P \times p er c a p i t a u se$
Substituting
$Q_{a} v g = 88, 920 (247) /1000 = 21, 963 m^{3} / d$
Peak day
$Q_{p} e ak = 2.1 \times 21, 963 = 46, 123 m^{3} / d$

Answer: P = 88,920 (geometric) vs 87,205 (arithmetic); Q_avg = 21,963 m³/d, Q_peak = 46,123 m³/d

Why the other options are there

15,490 (growth increment reported as population)
Q sized on the average day only

Reference: FE Reference Handbook — Environmental Engineering → Percent Growth

Example 7

Population projection and design water demand — Percent Growth (7)

A city of 138,278 grows at 0.9% per year. Project the population in 20 years by both geometric and arithmetic growth, then compute the average and peak day water demand at 345 L/capita/day.

Given

P₀ = 138,278
i = 0.9%/yr
n = 20 yr
Per capita use = 345 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
$P = 138278 (1 + 0.009)^{2} 0 = 165, 416 p eo pl e$
Formula (arithmetic)
$P = P_{0} (1 + in)$
Substituting
$P = 138278 (1 + 0.009 \times 20) = 163, 168 p eo pl e$
Formula
$Q_{a} v g = P \times p er c a p i t a u se$
Substituting
$Q_{a} v g = 165, 416 (345) /1000 = 57, 068 m^{3} / d$
Peak day
$Q_{p} e ak = 2.0 \times 57, 068 = 114, 137 m^{3} / d$

Answer: P = 165,416 (geometric) vs 163,168 (arithmetic); Q_avg = 57,068 m³/d, Q_peak = 114,137 m³/d

Why the other options are there

24,890 (growth increment reported as population)
Q sized on the average day only

Reference: FE Reference Handbook — Environmental Engineering → Percent Growth

Example 8

Population projection and design water demand — Percent Growth (8)

A city of 15,459 grows at 2.0% per year. Project the population in 14 years by both geometric and arithmetic growth, then compute the average and peak day water demand at 265 L/capita/day.

Given

P₀ = 15,459
i = 2.0%/yr
n = 14 yr
Per capita use = 265 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
$P = 15459 (1 + 0.020)^{1} 4 = 20, 398 p eo pl e$
Formula (arithmetic)
$P = P_{0} (1 + in)$
Substituting
$P = 15459 (1 + 0.020 \times 14) = 19, 788 p eo pl e$
Formula
$Q_{a} v g = P \times p er c a p i t a u se$
Substituting
$Q_{a} v g = 20, 398 (265) /1000 = 5, 405 m^{3} / d$
Peak day
$Q_{p} e ak = 2.5 \times 5, 405 = 13, 514 m^{3} / d$

Answer: P = 20,398 (geometric) vs 19,788 (arithmetic); Q_avg = 5,405 m³/d, Q_peak = 13,514 m³/d

Why the other options are there

4,329 (growth increment reported as population)
Q sized on the average day only

Reference: FE Reference Handbook — Environmental Engineering → Percent Growth

Example 9

Population projection and design water demand — Percent Growth (9)

A city of 108,849 grows at 1.0% per year. Project the population in 26 years by both geometric and arithmetic growth, then compute the average and peak day water demand at 380 L/capita/day.

Given

P₀ = 108,849
i = 1.0%/yr
n = 26 yr
Per capita use = 380 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
$P = 108849 (1 + 0.010)^{2} 6 = 140, 987 p eo pl e$
Formula (arithmetic)
$P = P_{0} (1 + in)$
Substituting
$P = 108849 (1 + 0.010 \times 26) = 137, 150 p eo pl e$
Formula
$Q_{a} v g = P \times p er c a p i t a u se$
Substituting
$Q_{a} v g = 140, 987 (380) /1000 = 53, 575 m^{3} / d$
Peak day
$Q_{p} e ak = 2.0 \times 53, 575 = 107, 150 m^{3} / d$

Answer: P = 140,987 (geometric) vs 137,150 (arithmetic); Q_avg = 53,575 m³/d, Q_peak = 107,150 m³/d

Why the other options are there

28,301 (growth increment reported as population)
Q sized on the average day only

Reference: FE Reference Handbook — Environmental Engineering → Percent Growth

Example 10

Population projection and design water demand — Percent Growth (10)

A city of 163,973 grows at 0.9% per year. Project the population in 18 years by both geometric and arithmetic growth, then compute the average and peak day water demand at 296 L/capita/day.

Given

P₀ = 163,973
i = 0.9%/yr
n = 18 yr
Per capita use = 296 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
$P = 163973 (1 + 0.009)^{1} 8 = 192, 670 p eo pl e$
Formula (arithmetic)
$P = P_{0} (1 + in)$
Substituting
$P = 163973 (1 + 0.009 \times 18) = 190, 537 p eo pl e$
Formula
$Q_{a} v g = P \times p er c a p i t a u se$
Substituting
$Q_{a} v g = 192, 670 (296) /1000 = 57, 030 m^{3} / d$
Peak day
$Q_{p} e ak = 2.0 \times 57, 030 = 114, 060 m^{3} / d$

Answer: P = 192,670 (geometric) vs 190,537 (arithmetic); Q_avg = 57,030 m³/d, Q_peak = 114,060 m³/d

Why the other options are there

26,564 (growth increment reported as population)
Q sized on the average day only

Reference: FE Reference Handbook — Environmental Engineering → Percent Growth

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

Percent Growth contains 5 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.

Percent Growth

Learning objectives

Lecture

Theory, developed

Notation used in this section

Handbook notes for this section

Core formulas for this FE topic

Pt=P0​(1+k)n

Pt=populationattimet

P0​=populationattimezero

k=growthrate

n=numberofperiods

Worked exam-style examples

Self-check

Chapter summary

Common traps in this section